Question

Light of certain wavelength $\lambda$ falls on a Potassium target. Photoelectrons released can be stopped completely by potential of 0.6 V. If frequency of incident light is increased by 10 %, stopping potential becomes 0.9 V, then

• A. Wavelength of original incident light is around 413 nm
• B. Work funciton of metal plate is 2.4 eV
• C. Work function of metal plate is 1.8 eV
• D. If $\lambda$ is increased by 10% new stopping potential would be 0.33 V

Answer 1

Answer: (A), (B), (D)

The correct options are
A Wavelength of original incident light is around 413 nm
B Work funciton of metal plate is 2.4 eV
D If $\lambda$ is increased by 10% new stopping potential would be 0.33 V

$V_o=\frac{hf}{e}-\frac{\varphi }{e}\Rightarrow 0.6=\frac{hf}{e}-\frac{\varphi }{e}...\left(1\right)\text{and}0.9=\frac{hf}{e}\times 1.1-\frac{\varphi }{e}...\left(2\right)$

$\left(2\right)-\left(1\right)\Rightarrow 0.3=0.1\times \frac{hf}{e}\Rightarrow \frac{hf}{e}=3$
By $\left(1\right),0.6=3-\frac{\varphi }{e}\Rightarrow \varphi =2.4eV$

$\frac{hf}{e}=3\Rightarrow \frac{hc}{\lambda e}=3\Rightarrow \lambda =\frac{hc}{3e}=\frac{1240}{3}nm=413nm$

$V_o=\frac{hc}{e\lambda }-\frac{\varphi }{e}=\frac{1240}{1.1\times 413}-2.4=0.33V$