NEB Solutions: Quantization of Energy

[2076 ‘A’]

Obtain the de Broglie wavelength of neutron of kinetic energy 150 eV. (mass of neutron = 1.675 x 10⁻²⁷ kg, h = 6.62 x 10⁻³⁴ Js).

Solution
de Broglie wavelength [latex](\lambda)[/latex] = ?
Kinetic energy (K.E.) = 150 eV = [latex]150\times 1.6\times 10^{-19} = 2.4\times 10^{-17}[/latex]
Mass of neutron (m) = [latex]1.675\times 10^{-27}[/latex] Kg
Planck’s Constant (h) = [latex]6.62\times 10^{-34}[/latex] Js

We have,

[latex]KE = \frac{1}{2}mv^2[/latex]

Or, [latex]v^2 = \frac{2KE}{m}[/latex]
Or, [latex]v = \sqrt{\frac{2KE}{m}} = \sqrt{\frac{2\times 2.4\times 10^{-17}}{1.675\times 10^{-27}}} = 1.693\times 10^5[/latex] m/s

Now, de Broglie wavelength [latex](\lambda) = \frac{h}{p} = \frac{h}{mv} = \frac{6.62\times 10^{-34}}{(1.675\times 10^{-27})(1.693\times 10^5)} = 2.334\times 10^{-12}[/latex] m.

An x – ray tube, operated at a DC potential difference of 10 KV, produces heat at the target at the rate of 720 watt. Assuming 0.5% of the incident electrons is converted into x – radiation, calculate the number of electrons striking per second at the target and velocity fo the incident electrons. (given e/m = 1.8 x 10¹⁷ Ckg⁻¹).

Solution
Given,
P.d. (V) = 10 KV = 10000 V
Production of heat (P) = 720 Watt.
Here, only 0.5% of the incident electrons is converted into x – radiation. This means, 99.5% of the incident electrons is converted into heat.
No. of electrons striking per second (n/t) = ?
Velocity of the incident electrons (v) = ?

We have,
Power (P) = VI
According to question,
Production of heat (P) = 99.5 % of IV
Or, [latex]720 = \frac{99.5}{100}\times IV[/latex]
Or, [latex]720 = 0.995 \times \frac{Q}{t}\times 10000[/latex]
Or, [latex]\frac{Q}{t} = \frac{720}{0.995\times 10000} = 7.236\times 10^{-2}[/latex]
Or, [latex]\frac{ne}{t} = 7.236\times 10^{-2}[/latex]
Or, [latex]\frac{n}{t} = \frac{7.236\times 10^{-2}}{1.6\times 10^{-19}} = 4.523\times 10^{17}[/latex] electrons/second.

Again, [latex]\frac{1}{2}mv^2 = eV[/latex]
So, [latex]v = \sqrt{\frac{2eV}{m}} = 5.93\times 10^7[/latex] m/s.

[2075 ‘B’]

A hydrogen atom is in ground state. What is the quantum number to which it will be excited absorbing a photon of energy 12.75 eV?