De Broglie Hypothesis UPSC Physics Optional Questions

1. The work function of a metal is $2 \times 10^{-19}$ joule. $ \\ $ (a) Find the threshold frequency for photoelectric emission $ \\ $ (b) Find the stopping potential if the metal is exposed to a frequency of $6 \times 10 ^ {14}$ hz.

(a) Threshold frequency is the minimum energy required to eject an electron from the surface of the metal. so $$ w_0 = h v_0 $$, where $w_0$ = work function, $v_0$ = threshold frequency and $h$ = planks constant. $ \\ $ $ \implies v_0 = \frac{w_0}{h} $ $ \\ $ $ \implies v_0 = \frac{2.5 \times 10 ^{-19}J}{6.6 \times 10^{-34} Js^{-1}} $ $ \\ $ $ \implies v_0 = 3.8 \times 10^{14} s^{-1}$ $ \\ $ (b) The energy of a photon is given by, $ \\ $ $ E = hv \implies E = 6.6 \times 10 ^ {-34} \times 6 \times 10^{14} = 3.9 \times 10^{-19} J \\$ this is the energy corresponding to the frequency given in the question. Now, the stopping potential is the maximum possible kinetic energy. So, $ \\ $ $ w_0 = 3.9 \times 10 ^ {-19} - 2 \times 10 ^ {-19} \\ $ $ w_0 = 1.9 \times 10 ^ {-19} J \\ $ $ w_0 = \frac{1.9 \times 10 ^ {-19}}{1.9 \times 10 ^ {-16}} eV \\ $ $ \implies w_0 = 1.2 eV $

2. An electron is confined to move between two rigid walls separated by $10^{-9}m$. Compute the De Broglie Wavelengths the first three allowed energy states of the electron and corresponding energies [PYQ]

Confined movement between two rigid walls resembles 1D potential box. Potential height of a rigid wall is infinity. Energy of particle confined in such a system is given by: $$ \boxed{E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}} $$ , where n = quantum number 1,2,3,... $\\$ m = mass of electron and L = length(width of 1D box) $\\$ Hence energy of first three allowed energy levels are $$ E_1 = \frac{ \pi^2 \hbar^2}{2mL^2}, E_2 = \frac{4 \pi^2 \hbar^2}{2mL^2}, E_3 = \frac{3 \pi^2 \hbar^2}{2mL^2} $$ And the corresponding wavelength (from De Broglie) would be: $\\$ using, $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE_n}}$, so $$ \lambda_1 = \frac{h}{\sqrt{2mE_1}}, \lambda_2 = \frac{h}{\sqrt{2mE_2}}, \lambda_3 = \frac{h}{\sqrt{2mE_3}} $$ $$ m=9.1 \times 10^{-31} Kg, \hbar = 1.054 \times 10^{-34} Js, L = 10^-9 m$$

3. An electron is confined to move between two rigid walls separated by $10^{-9}m$. Compute the De Broglie's wavelengths representing first three allowed energy states.

4. Derive Bohr's angular momentum quantization condition in Bohr's atomic model from concept of De Broglie waves

From De Broglie wavelength we know $$\tag{1} \lambda = \frac{h}{mv}$$, also $$ \tag{2} L=mvr $$, using both we have $$ \tag{3} L = r \times \frac{h}{\lambda} $$ $\\$ Now applying the standard wave concept in a circular orbit, $\\$ $\implies n \lambda = 2 \pi r \\$ $$ \tag{4} \implies \lambda = \frac{2 \pi r}{n} $$, combining these we get, $\implies L = r \times \frac{n}{2 \pi r} \times n$ $\\$ $$ \boxed{ L = \frac{nh}{2 \pi} } $$

5. What would be the wavelength of a quantum of radiant energy emitted if an electron is transmitted into radiation and went into one quantum?

The energy of a quantum is $E_1=hv=\frac{hc}{\lambda}.$ $$\\$$ The energy of an electron transmitted into radiation is $E_2=mc^2.$ $$\\$$ Here $E_1=E_2$ or,$\frac{hc}{\lambda}=mc^2$ or, $$ \lambda=\frac{h}{mc}=\frac{6.62\times{10}^{-34}}{9.1\times{10}^{-31}\times3\times{10}^8}=0.0244\mathrm{Å}$$

6. Estimate De Broglie wavelength of electron orbiting in first exicted state of hydrogen atom

The energy of n$^{th}$ state of $e^-$ in hydrogen atom is given by, $$ \boxed{ E = \frac{- 2 \pi^2 k^2 Z^2 e^4 m}{n^2 h^2} } $$ $ \implies E = \frac{-13.6eV}{n^2} \\$ $\\$ Energy of 1$^{st}$ excited state is when $n=2$: $\\$ $KE = -E \implies KE(E_k) = \frac{13.6eV}{4} \\$ $\lambda = \frac{h}{\sqrt{2mE_k}} \\$ $ \lambda = \frac{6.626 \times 10^{-34} \times 2} {\sqrt{2 \times 3.1 \times 10^{-31} \times 13.6 \times 1.6 \times 10^{-19}}}$ $\\$ $ \lambda = 6.6 Å $