Question

Radiation of wavelength 546 nm falls on a photo cathode and electrons with maximum kinetic energy of 0.18 eV are emitted. When radiation of wavelength 185 nm falls on the same surface, a (negative) stopping potential of 4.6 V has to be applied to the collector cathode to reduce the photoelectric current to zero. Then, the ratio h/e is:

• A. $6.6\times {10}^{-15}Js{C}^{-1}$
• B. $4.12\times {10}^{-15}Js{C}^{-1}$
• C. $6.6\times {10}^{-34}Js{C}^{-1}$
• D. $4.12\times {10}^{-34}Js{C}^{-1}$

Answer 1

Answer: (B)

$4.12\times {10}^{-15}Js{C}^{-1}$

Given:-

${\lambda }_1=546nm=546\times {10}^{-9}m$

$\left(K_{max}\right)=0.18eV=(0.18\times e)J$

${\lambda }_2=185nm=185\times {10}^{-9}m$

Stopping potential,$V=4.6V$(Nigative potential)

Solution:-

Let us assume the work function of the photo cathode is ${\varphi }_o$.

In the first case,

$\left(K_{max}\right)=\frac{hc}{{\lambda }_1}-{\varphi }_o$

$\therefore 0.18e=\frac{hc}{549\times {10}^{-9}}-{\varphi }_o⟶\left(1\right)$

In the second\quad case,

${\left(K_{max}\right)}_2=\frac{hc}{{\lambda }_2}-{\varphi }_o⟶\left(2\right)$

Here,

${\left(K_{max}\right)}_2=e\times Stoppingpotential)$
$[\because kineticenergyofelectronareconvertedtotheelectricpotentialenergy.]$

$\therefore {\left(K_{max}\right)}_2=e\times 4.6$

$usinginequation\left(2\right)4.6e=\frac{hc}{185\times {10}^{-9}}-{\varphi }_o⟶\left(3\right)$

Subtracting equation (1) from equation (3)

$\left(4.6e\right)-\left(0.18e\right)=hc[\frac{1}{185\times {10}^{-9}}-\frac{1}{546\times {10}^{-9}}]$

$4.42e=hc\times 0.00357\times {10}^9$

$\Rightarrow \frac{h}{e}=\frac{4.42}{3\times {10}^8\times 0.00357\times {10}^9}$

$\frac{h}{e}=4.127\times {10}^{-15}Js{C}^{-1}$