[2075]
A clean nickel surface of work function 5.1 eV is exposed to light of wavelength 235 nm. What is the maximum speed of the photoelectrons emitted from its surface?
Solution
Given,
Work function [latex](\phi) = 5.1\ eV[/latex]
= [latex]5.1\times 1.6\times 10^{-19} = 8.16\times 10^{-19}[/latex] J
Wavelength [latex](\lambda)[/latex] = 235 nm = [latex]235\times 10^{-9}[/latex]
Speed (v) = ?
We have,
[latex]E = KE + \phi[/latex]
Or, [latex]hf = KE + \phi[/latex]
Or, [latex]h\frac{c}{\lambda} – \phi = KE[/latex]
Or, [latex]KE = (6.62\times 10^{-34})\frac{3\times 10^8}{235\times 10^{-9}} – (8.16\times 10^{-19})[/latex]
Or, [latex]KE = 2.9106\times 10^{-20}[/latex]
Now, [latex]KE = \frac{1}{2}mv^2[/latex]
Or, [latex]v = \sqrt{\frac{2KE}{m}} = \sqrt{\frac{2\times 2.9106\times 10^{-20}}{9.1\times 10^{-31}}} = 2.529\times 10^5[/latex] m/s.
[2075 ‘B’/2072 ‘S’]
Sodium has a work function of 2eV. Calculate the maximum energy and speed of the emitted electrons when sodium is illuminated by a radiation of 150 nm. What is the threshold frequency of radiation for which electrons are emitted from sodium surface?
Solution
Given,
Work function [latex](\phi)[/latex] = 2 eV = [latex]2\times 1.6\times 10^{-19} = 3.2\times 10^{-19}[/latex]
Maximum energy of electrons (K.E.) = ?
Speed of electrons (v) = ?
Wavelength of radiation [latex](\lambda)[/latex] = 150 nm = [latex]150\times 10^{-9}[/latex] m
Threshold frequency (f₀) = ?
We know that,
[latex]E = K.E. + \phi[/latex]
Or, [latex]K.E. = E – \phi[/latex]
[latex]= hf – \phi[/latex]
[latex]= h\frac{c}{\lambda} – \phi[/latex]
[latex]= (6.62\times 10^{-34})\frac{3\times 10^8}{(150\times 10^{-9})} – (3.2\times 10^{-19})[/latex]
[latex]= 1.004\times 10^{-18}[/latex] J
Now, [latex]K.E. = \frac{1}{2}mv^2[/latex]
Or, [latex]v^2 = \frac{2K.E.}{m}[/latex]
Or, [latex]v = \sqrt{\frac{2K.E.}{m}} = \sqrt{\frac{1.004\times 10^{-18}}{(9.1\times 10^{-31})}} = 1.050\times 10^6[/latex] m/s
Again, [latex]Work\ function\ (\phi) = hf_o[/latex]
Or, [latex]3.2\times 10^{-19} = (6.62\times 10^{-34})f_o[/latex]
Or, [latex]f_o = \frac{3.2\times 10^{-19}}{6.62\times 10^{-34}} = 4.834\times 10^{14}[/latex] Hz.
[2074 ‘B’]
Radiations of wavelength 5400 Å fall on a metal whose work function is 1.9 eV. Find the energy of the photoelectrons emitted and their stopping potential. (Planck’s Constant = 6.62 x 10⁻³⁴ Js).
Solution
Given,
Wavelength of radiation [latex](\lambda)[/latex] = 5400 Å = [latex]5400\times 10^{-10}[/latex] m
Work function [latex](\phi)[/latex] = 1.9 eV = [latex]1.9\times 1.6\times 10^{-19} = 3.04\times 10^{-19}[/latex] J
Energy of photoelectrons emitted (K.E.) = ?
Stopping Potential (V₀) = ?
Planck’s constant (h) = [latex]6.62\times 10^{-34}[/latex] Js
We know that,
[latex]E = K.E. + \phi[/latex]
Or, [latex]K.E. = E – \phi[/latex]
[latex]= hf – \phi[/latex]
[latex]= h\frac{c}{\lambda} – \phi[/latex]
[latex]= (6.62\times 10^{-34})\frac{3\times 10^8}{5400\times 10^{-10}} – (3.04\times 10^{-19})[/latex]
[latex]= 6.378\times 10^{-20}[/latex] J
[latex]= \frac{6.378\times 10^{-20}}{1.6\times 10^{-19}},eV = 0.3986[/latex] eV
Again, we know,
[latex]K.E. = eV_o[/latex]
Or, [latex]V_o = \frac{K.E.}{e} = \frac{0.3986,eV}{e} = 0.3986\ V[/latex].
[2073 ‘S’]
Light of frequency 5 x 10¹⁴ Hz liberates electrons with energy 2.3 x 10⁻¹⁹ J from a certain metallic surface. What is the wavelength of ultraviolet light which liberates electrons of energy 8.93 x 10⁻¹⁹ J from the same surface? (Given h = 6.62 x 10⁻³⁴ Js, c = 3 x 10⁸ m/s).
Solution
Given,
Frequency of light (f₁) = [latex]5\times 10^{14}[/latex] Hz
Energy of electron (K.E.₁) = [latex]2.3\times 10^{-19}[/latex] J
Wavelength of light [latex](\lambda)[/latex] = ?
Energy of electron (K.E.₂) = [latex]8.93\times 10^{-19}[/latex] J
Planck’s constant (h) = [latex]6.62\times 10^{-34}[/latex] Js
Speed of light (c) = [latex]3\times 10^8[/latex] m/s
We know that,
[latex]E = K.E. + \phi[/latex]
So, for first case,
[latex]E_1 = K.E._1 + \phi[/latex] ……………………. (i)
Also, for second case,
[latex]E_2 = K.E._2 + \phi[/latex] ……………………….. (ii)
Now, subtracting eqn. (i) from eqn. (ii), we get,
[latex]E_2 – E_1 = K.E._2 + \phi – K.E._1 – \phi[/latex]
Or, [latex]E_2 – E_1 = K.E._2 – K.E._1[/latex]
Or, [latex]E_2 – E_1 = (K.E._2 – K.E._1)[/latex]
Or, [latex]hf_2 – hf_1 = (8.93\times 10^{-19} – 2.3\times 10^{-19})[/latex]
Or, [latex]h(f_2 – f_1) = (6.63\times 10^{-19})[/latex]
Or, [latex]f_2 – f_1 = \frac{6.63\times 10^{-19}}{h}[/latex]
Or, [latex]f_2 = \frac{6.63\times 10^{-19}}{6.62\times 10^{-34}} + f_1[/latex]
Or, [latex]\frac{c}{\lambda_2} = 1.002\times 10^{15} + 5\times 10^{14}[/latex]
Or, [latex]\frac{c}{\lambda_2} = 1.502\times 10^{15}[/latex]
Or, [latex]\lambda_2 = \frac{c}{1.502\times 10^{15}} = \frac{3\times 10^8}{1.502\times 10^{15}} = 1.997\times 10^{-7}[/latex] m.
[2073 ‘D’]
400 nm wavelength of light falls on a photo sensitive material of work function 2.3 eV. Compute the maximum energy of photoelectrons.
Solution
Given,
Wavelength of light [latex](\lambda)[/latex] = 400 nm = [latex]400\times 10^{-9}[/latex] m
Work function [latex](\phi)[/latex] = 2.3 eV = [latex](2.3\times 1.6\times 10^{-19}) = 3.68\times 10^{-19}[/latex] J
Maximum energy of photoelectrons [latex](K.E._{max})[/latex] = ?
We know that,
[latex]E = K.E._{max} + \phi[/latex]
Or, [latex]K.E._{max} = E – \phi[/latex]
[latex]= hf – (3.68\times 10^{-19})[/latex]
[latex]= \frac{hc}{\lambda} – (3.68\times 10^{-19})[/latex]
[latex]= \frac{(6.62\times 10^{-34})\times(3\times 10^8)}{(400\times 10^{-9})} – (3.68\times 10^{-19})[/latex]
[latex]= 1.285\times 10^{-19}[/latex]
[latex]= \frac{1.285\times 10^{-19}}{(1.6\times 10^{-19})},eV = 0.803\ eV[/latex].
[2072 ‘S’]
Sodium has a work function of 2eV. Calculate the maximum energy and speed of the emitted electrons when sodium is illuminated by radiation of wavelength 150 nm. (Given mass of electron = 9.1 x 10⁻³¹ kg).
Solution
Given,
Work function [latex](\phi)[/latex] = 2eV = [latex]2\times 1.6\times 10^{-19} = 3.2\times 10^{-19}[/latex] J
Maximum energy of the emitted electrons (KE) = ?
Maximum speed (v) = ?
Wavelength of radiation [latex](\lambda)[/latex] = 150 nm = [latex]150\times 10^{-9}[/latex] m
Mass of electron (m) = [latex]9.1\times 10^{-31}[/latex] kg
We know that,
[latex]E = K.E. + \phi[/latex]
Or, [latex]K.E. = E – \phi[/latex]
[latex]= hf – \phi[/latex]
[latex]= h\frac{c}{\lambda} – \phi[/latex]
[latex]= (6.62\times 10^{-34})\frac{3\times 10^8}{150\times 10^{-9}} – (3.2\times 10^{-19})[/latex]
[latex]= 1.004\times 10^{-18}[/latex] J
Again,
[latex]K.E. = \frac{1}{2}mv^2[/latex]
Or, [latex]v^2 = \frac{2K.E.}{m}[/latex]
Or, [latex]v = \sqrt{\frac{2K.E.}{m}} = \sqrt{\frac{2\times(1.004\times 10^{-18})}{9.1\times 10^{-31}}} = 1.485\times 10^6[/latex] m/s.
[2072 ‘C’]
The work function for the surface of aluminium is 4.2 eV. How much potential difference will be required to stop the emission of maximum energy electrons emitted by light of wavelength 2000 Å? (Planck’s constant = 6.6 x 10⁻³⁴ Js).
Solution
Given,
Work function for the aluminium surface [latex](\phi)[/latex] = 4.2 eV = [latex]4.2\times 1.6\times 10^{-19}[/latex] J = [latex]6.72\times 10^{-19}[/latex] J
Stopping Potential (V₀) = ?
Wavelength of light [latex](\lambda)[/latex] = 2000 Å = [latex]2000\times 10^{-10}[/latex] m
Planck’s constant (h) = [latex]6.6\times 10^{-34}[/latex] Js
We know that,
[latex]E = K.E. + \phi[/latex]
Or, [latex]K.E. = E – \phi[/latex]
Or, [latex]eV_o = hf – \phi[/latex]
Or, [latex]V_o = \frac{hf-\phi}{e} = \frac{h\frac{c}{\lambda}-\phi}{e} = \frac{(6.6\times 10^{-34})\times\frac{3\times 10^8}{2000\times 10^{-10}} – (6.72\times 10^{-19})}{(1.6\times 10^{-19})} = 1.987[/latex] V.
Light of wavelength 5 x 10⁻⁷ m falls on a sensitive metal plate with photo electric work function 1.90 eV. Find Kinetic energy of the photoelectrons emitted and stopping potential. (Given h = 6.62 x 10⁻³⁴ Js).
Solution
Given,
Wavelength of light [latex](\lambda)[/latex] = [latex]5\times 10^{-7}[/latex] m
Work function [latex](\phi)[/latex] = 1.90 eV
= [latex]1.9\times 1.6\times 10^{-19} = 3.040\times 10^{-19}[/latex] J
Kinetic energy of photoelectrons (K.E.) = ?
Stopping potential (V₀) = ?
We know that,
[latex]E = K.E. + \phi[/latex]
Or, [latex]K.E. = E – \phi[/latex]
[latex]= hf – \phi[/latex]
[latex]= \frac{hc}{\lambda} – \phi[/latex]
[latex]= \frac{(6.62\times 10^{-34})\times(3\times 10^8)}{(5\times 10^{-7})} – (3.04\times 10^{-19})[/latex]
[latex]= 9.32\times 10^{-20}[/latex] J
[latex]= \frac{9.32\times 10^{-20}}{1.6\times 10^{-19}} = 0.5825\ eV[/latex]
Now, We also have,
[latex]K.E. = eV_o[/latex]
Or, [latex]V_o = \frac{K.E.}{e} = \frac{0.5825,eV}{e} = 0.5825\ V[/latex].
Ultraviolet light of wavelength 3.6 x 10⁻⁷ m is made to fall on a smooth surface of potassium. The work function of potassium is 2 eV. Determine maximum energy of emitted photo electrons and stopping potential.
Solution
Given,
Wavelength of light [latex](\lambda)[/latex] = [latex]3.6\times 10^{-7}[/latex] m
Work function [latex](\phi)[/latex] = 2 eV
= [latex]2\times 1.6\times 10^{-19}[/latex] J = [latex]3.2\times 10^{-19}[/latex] J
Maximum energy of photoelectrons [latex](K.E._{max})[/latex] = ?
Also, stopping potential (V₀) = ?
We know that,
[latex]E = K.E._{max} + \phi[/latex]
Or, [latex]K.E._{max} = E – \phi[/latex]
[latex]= hf – \phi[/latex]
[latex]= \frac{hc}{\lambda} – \phi[/latex]
[latex]= \frac{(6.62\times 10^{-34})\times(3\times 10^8)}{(3.6\times 10^{-7})} – (3.2\times 10^{-19})[/latex]
[latex]= 2.317\times 10^{-19}[/latex] J
[latex]= \frac{2.317\times 10^{-19}}{1.6\times 10^{-19}} = 1.448\ eV[/latex]
Also, we have,
[latex]K.E._{max} = eV_o[/latex]
Or, [latex]V_o = \frac{K.E._{max}}{e} = \frac{1.448,eV}{e} = 1.448[/latex] V
Electrons with maximum kinetic energy of 3 eV are ejected from a metal surface by ultra-violet radiation of wavelength 1.5 x 10⁻⁷ m. Determine work function, threshold wavelength and the stopping potential for the metal. (Planck’s constant h = 6.62 x 10⁻³⁴ Js).
Solution
Given,
Maximum KE [latex](K.E._{max})[/latex] = 3 eV = [latex]3\times 1.6\times 10^{-19} = 4.8\times 10^{-19}[/latex] J
Wavelength of radiation [latex](\lambda)[/latex] = [latex]1.5\times 10^{-7}[/latex] m
Work function [latex](\phi)[/latex] = ?
Threshold wavelength [latex](\lambda_o)[/latex] = ?
Stopping potential (V₀) = ?
Planck’s constant (h) = [latex]6.62\times 10^{-34}[/latex] Js
We know that,
[latex]E = K.E._{max} + \phi[/latex]
Or,
[latex]\phi = E – K.E._{max}[/latex]
[latex]= hf – KE_{max}[/latex]
[latex]= \frac{hc}{\lambda} – KE_{max}[/latex]
[latex]= \frac{(6.62\times 10^{-34})\times(3\times 10^8)}{(1.5\times 10^{-7})} – (4.8\times 10^{-19})[/latex]
[latex]= 8.44\times 10^{-19}[/latex] J
Now,
[latex]\phi = hf_o[/latex]
Or, [latex]\phi = \frac{hc}{\lambda_o}[/latex]
Or, [latex]\lambda_o = \frac{hc}{\phi} = \frac{(6.62\times 10^{-34})\times(3\times 10^8)}{8.44\times 10^{-19}} = 2.353\times 10^{-7}[/latex] m
Also, we have,
[latex]K.E._{max} = eV_o[/latex]
Or, [latex]V_o = \frac{K.E._{max}}{e} = \frac{4.8\times 10^{-19}}{1.6\times 10^{-19}} = 3[/latex] V
The photoelectric threshold wavelength of a tungsten surface is 272 nm. Calculate the maximum velocity of the electrons ejected from this tungsten surface by ultraviolet radiation of frequency 1.45 x 10¹⁵ Hz. (h = 6.62 x 10⁻³⁴ Js, c = 3 x 10⁸ m/s, mₑ = 9.1 x 10⁻³¹ kg).
Solution
Given,
Threshold wavelength [latex](\lambda_o)[/latex] = 272 nm = [latex]272\times 10^{-9}[/latex] m
Maximum velocity [latex](v_{max})[/latex] = ?
Frequency of radiation (f) = [latex]1.45\times 10^{15}[/latex] Hz
Planck’s constant (h) = [latex]6.62\times 10^{-34}[/latex] Js
Velocity of light (c) = [latex]3\times 10^8[/latex] m/s
Mass of electron (mₑ) = [latex]9.1\times 10^{-31}[/latex] kg
We know that,
[latex]E = K.E. + \phi[/latex]
Or, [latex]K.E. = E – \phi[/latex]
[latex]= hf – hf_o[/latex]
[latex]= hf – \frac{h\lambda_o}{c}[/latex]
[latex]= (6.62\times 10^{-34})(1.45\times 10^{15}) – \frac{(6.62\times 10^{-34})\times(3\times 10^8)}{272\times 10^{-9}}[/latex]
[latex]= 2.298\times 10^{-19}[/latex] J
Again,
[latex]K.E. = \frac{1}{2}mv_{max}^2[/latex]
Or, [latex]v_{max}^2 = \frac{2K.E.}{m}[/latex]
Or, [latex]v_{max} = \sqrt{\frac{2K.E.}{m}} = \sqrt{\frac{2\times(2.298\times 10^{-19})}{9.1\times 10^{-31}}} = 7.107\times 10^5[/latex] m/s.
When ultraviolet light with wavelength of 400 nm falls on a certain metal surface, the maximum kinetic energy of the emitted electrons is 1.10 eV. What is the maximum kinetic energy of the photoelectrons when light of wavelength 300 nm falls on the same surface?
Solution
Given,
Wavelength of light [latex](\lambda_1)[/latex] = 400 nm = [latex]400\times 10^{-9}[/latex] m
Maximum K.E. (K.E.₁) = 1.1 eV = [latex]1.1\times 1.6\times 10^{-19} = 1.76\times 10^{-19}[/latex] J
Maximum K.E. (K.E.₂) = ?
Wavelength of light [latex](\lambda_2)[/latex] = 300 nm = [latex]300\times 10^{-9}[/latex] m
We know that,
[latex]E = K.E. + \phi[/latex]
Or, [latex]h\frac{c}{\lambda} = K.E. + \phi[/latex]
For 1st case,
[latex]\frac{hc}{\lambda_1} = K.E._1 + \phi[/latex] ……………………. (i)
[latex]For\ 2^{nd} case,[/latex]
[latex]\frac{hc}{\lambda_2} = K.E._2 + \phi[/latex] ……………………. (ii)
Subtracting eqn. (i) from (ii), we get,
[latex]\frac{hc}{\lambda_2} – \frac{hc}{\lambda_1} = K.E._2 – K.E._1[/latex]
Or, [latex]hc\left(\frac{1}{\lambda_2} – \frac{1}{\lambda_1}\right) + K.E._1 = K.E._2[/latex]
Or, [latex]K.E._2 = (6.62\times 10^{-34})\times(3\times 10^8)\left(\frac{1}{300\times 10^{-9}} – \frac{1}{400\times 10^{-9}}\right) + (1.76\times 10^{-19})[/latex]
[latex]= 3.415\times 10^{-19}[/latex] J
[latex]= \frac{3.415\times 10^{-19}}{1.6\times 10^{-19}} = 2.134[/latex] eV.
What will be the change in the stopping potential for photoelectrons emitted from a source if the wavelength of incident light is reduced from 400 nm to 360 nm?
Solution
Given,
Change in stopping potential [latex](\Delta V_o)[/latex] = ?
Wavelength of incident light in 1st case [latex](\lambda_1)[/latex] = 400 nm = [latex]400\times 10^{-9}[/latex] m
Wavelength of incident light in 2nd case [latex](\lambda_2)[/latex] = 360 nm = [latex]360\times 10^{-9}[/latex] m
We know that,
[latex]E = K.E. + \phi[/latex]
Or, [latex]hf = K.E. + \phi[/latex]
Or, [latex]h\frac{c}{\lambda} = eV_o + \phi[/latex]
For 1st case,
[latex]\frac{hc}{\lambda_1} = eV_{o1} + \phi[/latex] ………………….. (i)
For 2nd case,
[latex]\frac{hc}{\lambda_2} = eV_{o2} + \phi[/latex] ……………………… (ii)
On subtracting,
[latex]\frac{hc}{\lambda_2} – \frac{hc}{\lambda_1} = eV_{o2} – eV_{o1}[/latex]
Or, [latex]hc\left(\frac{1}{\lambda_2} – \frac{1}{\lambda_1}\right) = e(V_{o2} – V_{o1})[/latex]
Or, [latex]hc\left(\frac{1}{\lambda_2} – \frac{1}{\lambda_1}\right) = e\Delta V_o[/latex]
Or, [latex]\Delta V_o = \frac{hc}{e}\left(\frac{1}{\lambda_2} – \frac{1}{\lambda_1}\right)[/latex]
[latex]= \frac{(6.62\times 10^{-34})\times(3\times 10^8)}{1.6\times 10^{-19}}\left(\frac{1}{360\times 10^{-9}} – \frac{1}{400\times 10^{-9}}\right)[/latex]
[latex]= 0.345[/latex] V.
The maximum energy of photoelectrons emitted from a metal plate 1.2 eV. If the threshold wavelength is 2.48 x 10⁻⁹ m. Calculate the wavelength of incident light. (Given plank’s constant is 6.62 x 10⁻³⁴ Js).
Solution
Given,
Maximum energy of photoelectron (K.E.) = 1.2 eV = [latex]1.2\times 1.6\times 10^{-19} = 1.92\times 10^{-19}[/latex] J
Threshold wavelength [latex](\lambda_o)[/latex] = [latex]2.48\times 10^{-9}[/latex] m
Wavelength of incident light [latex](\lambda)[/latex] = ?
Planck’s constant (h) = [latex]6.62\times 10^{-34}[/latex] Js
We know that,
[latex]E = K.E. + \phi[/latex]
Or, [latex]hf = K.E. + hf_o[/latex]
Or, [latex]h\frac{c}{\lambda} = K.E. + \frac{hc}{\lambda_o}[/latex]
Or, [latex]\frac{(6.62\times 10^{-34})\times(3\times 10^8)}{\lambda} = 1.92\times 10^{-19} + \frac{(6.62\times 10^{-34})\times(3\times 10^8)}{(2.48\times 10^{-9})}[/latex]
Or, [latex]\frac{1.986\times 10^{-25}}{\lambda} = 8.027\times 10^{-17}[/latex]
Or, [latex]\lambda = 2.474\times 10^{-9}[/latex] m.
Light of wavelength 6000 Å falls on a photosensitive plate of work function 1.9 eV. Find the speed of the photo electrons emitted. (h = 6.62 x 10⁻³⁴ Js and mₑ = 9.1 x 10⁻³¹ Kg).
Solution
Given,
Wavelength of light [latex](\lambda)[/latex] = 6000 Å = [latex]6000\times 10^{-10}[/latex] m
Work function [latex](\phi)[/latex] = 1.9 eV = [latex]1.9\times 1.6\times 10^{-19} = 3.04\times 10^{-19}[/latex] J
Speed of photoelectrons (v) = ?
Planck’s constant (h) = [latex]6.62\times 10^{-34}[/latex] Js
Mass of electron (m) = [latex]9.1\times 10^{-31}[/latex] kg
We know that,
[latex]E = K.E. + \phi[/latex]
Or, [latex]K.E. = E – \phi[/latex]
[latex]= h\frac{c}{\lambda} – \phi[/latex]
[latex]= (6.62\times 10^{-34})\frac{3\times 10^8}{6000\times 10^{-10}} – 3.04\times 10^{-19}[/latex]
[latex]= 2.7\times 10^{-20}[/latex] J
Again,
[latex]K.E. = \frac{1}{2}mv^2[/latex]
Or, [latex]v^2 = \frac{2K.E.}{m}[/latex]
Or, [latex]v = \sqrt{\frac{2K.E.}{m}} = \sqrt{\frac{2\times(2.7\times 10^{-20})}{9.1\times 10^{-31}}} = 2.436\times 10^5[/latex] m/s.
When a light of frequency 5.4 x 10¹⁴ Hz is incident on a metal surface, the maximum energy of the electrons emitted is 1.2 x 10⁻¹⁹ J. If the same surface is illuminated with light of frequency 6.6 x 10¹⁴ Hz, the maximum energy of the electrons is 2 x 10⁻¹⁹ J. Find the value of Planck’s constant.
Solution
Given,
Frequency of light (f₁) = [latex]5.4\times 10^{14}[/latex] Hz
Maximum energy of electrons (K.E.₁) = [latex]1.2\times 10^{-19}[/latex] J
Frequency of light in 2nd case (f₂) = [latex]6.6\times 10^{14}[/latex] Hz
Maximum energy of electrons in 2nd case (K.E.₂) = [latex]2\times 10^{-19}[/latex] J
Planck’s constant (h) = ?
We know that,
[latex]E = K.E. + \phi[/latex]
Or, [latex]hf = K.E. + \phi[/latex]
For 1st case,
[latex]hf_1 = K.E._1 + \phi[/latex] ………………….. (i)
For 2nd case,
[latex]hf_2 = K.E._2 + \phi[/latex] ……………………… (ii)
On subtracting, we get,
[latex]hf_2 – hf_1 = K.E._2 – K.E._1[/latex]
Or, [latex]h(f_2 – f_1) = K.E._2 – K.E._1[/latex]
Or, [latex]h = \frac{K.E._2 – K.E._1}{f_2 – f_1} = \frac{(2\times 10^{-19}) – (1.2\times 10^{-19})}{(6.6\times 10^{14}) – (5.4\times 10^{14})} = 6.667\times 10^{-34}[/latex] Js.
Radiation of 3000 Å falls on a photo sensitive material of work function 3 eV and photoelectrons are ejected. Find the maximum kinetic energy of photoelectrons.
Solution
Given,
Wavelength of radiation [latex](\lambda)[/latex] = 3000 Å = [latex]3000\times 10^{-10}[/latex] m
Work function [latex](\phi)[/latex] = 3 eV = [latex]3\times 1.6\times 10^{-19} = 4.8\times 10^{-19}[/latex] J
Maximum K.E. of photoelectrons [latex](K.E._{max})[/latex] = ?
We know that,
[latex]E = K.E._{max} + \phi[/latex]
Or, [latex]K.E._{max} = E – \phi[/latex]
[latex]= h\frac{c}{\lambda} – \phi[/latex]
[latex]= (6.62\times 10^{-34})\frac{3\times 10^8}{3000\times 10^{-10}} – 4.8\times 10^{-19}[/latex]
[latex]= 1.82\times 10^{-19}[/latex] J.
A UV light of 400 nm strikes a cesium surface of work function 1.9 eV. Find the velocity of electron emitted from the cesium surface. (mₑ = 9.1 x 10⁻³¹ kg, c = 3 x 10⁸ m/s, h = 6.62 x 10⁻³⁴ Js).
Solution
Given,
Wavelength of UV light [latex](\lambda)[/latex] = 400 nm = [latex]400\times 10^{-9}[/latex] m
Work function [latex](\phi)[/latex] = 1.9 eV = [latex]1.9\times 1.6\times 10^{-19} = 3.04\times 10^{-19}[/latex] J
Velocity of electron (v) = ?
Mass of electron (m) = [latex]9.1\times 10^{-31}[/latex] kg
Velocity of light (c) = [latex]3\times 10^8[/latex] m/s
Planck’s constant (h) = [latex]6.62\times 10^{-34}[/latex] Js
We know that,
[latex]E = K.E. + \phi[/latex]
Or, [latex]K.E. = E – \phi[/latex]
[latex]= h\frac{c}{\lambda} – \phi[/latex]
[latex]= (6.62\times 10^{-34})\frac{3\times 10^8}{400\times 10^{-9}} – 3.04\times 10^{-19}[/latex]
[latex]= 1.925\times 10^{-19}[/latex] J
Again,
[latex]K.E. = \frac{1}{2}mv^2[/latex]
Or, [latex]v^2 = \frac{2K.E.}{m}[/latex]
Or, [latex]v = \sqrt{\frac{2K.E.}{m}} = \sqrt{\frac{2\times(1.925\times 10^{-19})}{9.1\times 10^{-31}}} = 6.504\times 10^5[/latex] m/s.
The photoelectric work function of potassium is 2eV and the surface is illuminated with radiation of wavelength 350 nm. What potential difference have to be applied between a potassium surface and the collecting electrode in order just to prevent collection of electrons? What would be the kinetic energy of the electrons?
Solution
Given,
Work function [latex](\phi)[/latex] = 2 eV = [latex]2\times 1.6\times 10^{-19} = 3.2\times 10^{-19}[/latex] J
Wavelength of radiation [latex](\lambda)[/latex] = 350 nm = [latex]350\times 10^{-9}[/latex] m
Potential difference (V₀) = ?
Kinetic energy of electrons (K.E.) = ?
We know that,
[latex]E = K.E. + \phi[/latex]
Or, [latex]K.E. = E – \phi[/latex]
[latex]= h\frac{c}{\lambda} – \phi[/latex]
[latex]= (6.62\times 10^{-34})\frac{3\times 10^8}{350\times 10^{-9}} – 3.2\times 10^{-19}[/latex]
[latex]= 2.474\times 10^{-19}[/latex] J
Again, K.E. required to prevent electrons from colliding,
[latex]K.E. = eV_o[/latex]
Or, [latex]V_o = \frac{K.E.}{e} = \frac{2.474\times 10^{-19}}{1.6\times 10^{-19}} = 1.546[/latex] V.
The maximum kinetic energy of the electrons emitted from a metallic surface is 1.6 x 10⁻¹⁹ J when the frequency of the radiation is 7.5 x 10¹⁴ Hz. Calculate the minimum frequency of the radiation for which electrons will be emitted. Assume that h = 6.6 x 10⁻³⁴ Js.
Solution
Given,
Maximum Kinetic energy [latex](K.E._{max})[/latex] = [latex]1.6\times 10^{-19}[/latex] J
Frequency of radiation (f) = [latex]7.5\times 10^{14}[/latex] Hz
Minimum frequency (f₀) = ?
Planck’s constant (h) = [latex]6.6\times 10^{-34}[/latex] Js
We know that,
[latex]E = K.E._{max} + \phi[/latex]
Or, [latex]hf = K.E._{max} + hf_o[/latex]
Or, [latex]hf_o = hf – K.E._{max}[/latex]
Or, [latex]f_o = \frac{hf – K.E._{max}}{h} = \frac{(6.6\times 10^{-34})\times(7.5\times 10^{14}) – (1.6\times 10^{-19})}{(6.6\times 10^{-34})} = 5.076\times 10^{14}[/latex] Hz