Question

In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength (λ) of incident light at the corresponding stopping potential (${V_0}$) are given below:

$\lambda \left( {\mu m} \right)$	${V_0}$ (volts)
0.3	2.0
0.4	1.0
0.5	0.4

Given that $c = 3 \times {10^8}{\text{ m/s}}$ and $e = 1.6 \times {10^{ - 19}}C$, Planck's constant (in units of Joule-seconds) from such an experiment is:
A. $6.0 \times {10^{ - 34}}$
B. $6.4 \times {10^{ - 34}}$
C. $6.6 \times {10^{ - 34}}$
D. $6.8 \times {10^{ - 34}}$

Answer 1

In this question, we need to determine Planck's constant through the data given carried out by historical experiment. For this, we will use the relation between the total energy of emission of the electrons, work done and the kinetic energy of the emitted electrons.

Complete step by step answer:
The total energy of the emitted electrons will be the sum of the work done to emit those electrons and the maximum kinetic energy that they achieve during their transition. Mathematically, $E = W + KE$.
Let the wavelength of the emitted electrons be $\lambda$ when a potential of ${V_0}$ has been applied across the anode and the cathode. So, the equation $E = W + KE$ will be reduced to

$$E = W + KE \\\ \dfrac{{hc}}{\lambda } = \dfrac{{hc}}{{{\lambda _0}}} + e{V_0} - - - - (i) \\\$$

Now, substituting the values given in the table in the equation (i).

$\lambda \left( {\mu m} \right)$	${V_0}$ (volts)
0.3	2.0
0.4	1.0
0.5	0.4

$$\dfrac{{hc}}{\lambda } = \dfrac{{hc}}{{{\lambda _0}}} + e{V_0} \\\ \dfrac{{hc}}{{0.3 \times {{10}^{ - 6}}}} = \dfrac{{hc}}{{{\lambda _0}}} + e \times 2 - - - - - (ii) \\\$$

Again,

$$\dfrac{{hc}}{\lambda } = \dfrac{{hc}}{{{\lambda _0}}} + e{V_0} \\\ \dfrac{{hc}}{{0.4 \times {{10}^{ - 6}}}} = \dfrac{{hc}}{{{\lambda _0}}} + e \times 1 - - - - - (iii) \\\$$

Subtracting the equation (iii) from the equation (ii) we get

$$\dfrac{{hc}}{{0.3 \times {{10}^{ - 6}}}} - \dfrac{{hc}}{{0.4 \times {{10}^{ - 6}}}} = \left( {\dfrac{{hc}}{{{\lambda _0}}} + e \times 2} \right) - \left( {\dfrac{{hc}}{{{\lambda _0}}} + e \times 1} \right) \\\ hc\left( {\dfrac{1}{{0.3 \times {{10}^{ - 6}}}} - \dfrac{1}{{0.4 \times {{10}^{ - 6}}}}} \right) = e - - - - (iv) \\\$$

Substituting the values of $c = 3 \times {10^8}{\text{ m/s}}$ and $e = 1.6 \times {10^{ - 19}}C$ in the equation (iv), we get

hc\left( {\dfrac{1}{{0.3 \times {{10}^{ - 6}}}} - \dfrac{1}{{0.4 \times {{10}^{ - 6}}}}} \right) = e \\\ \Rightarrow h \times 3 \times {10^8} \times {10^6}\left( {\dfrac{{0.4 - 0.3}}{{0.3 \times 0.4}}} \right) = 1.6 \times {10^{ - 19}} \\\ \Rightarrow h = \dfrac{{1.6 \times {{10}^{ - 19}}}}{{3 \times {{10}^{8 + 6}}}}\left( {\dfrac{{0.12}}{{0.1}}} \right) \\\ \Rightarrow h = 0.53 \times {10^{ - 19 - 8 - 6}}\left( {1.2} \right) \\\ \Rightarrow h = 0.64 \times {10^{ - 33}} \\\ $$ $\therefore h = 6.4 \times {10^{ - 34}}Js \\\ $ **Hence, the value of the Planck's constant on performing the experiment is $6.4 \times {10^{ - 34}}Js$. So, Option B is correct.** **Note:** The value of the wavelength in the given table is in micrometers so, we have to convert it first in meters and then substitute it in the desired equation. Moreover, the actual value of the Planck's constant is given as $6.626 \times {10^{ - 34}}Js$ which is comparable with our experimental value, so the result is correct.