Question

Given that a photon of light of wavelength $10,000Å$ has an energy equal to $1.23eV$ . When light of wavelength $5000Å$ and intensity $I_0$ falls on a photoelectric cell, the surface current is $0.40\times 10^{-6}A$ and the stopping potential is $1.36V$ , then the work function is

• A. $0.43eV$
• B. $0.55eV$
• C. $1.10eV$
• D. $1.53eV$

Answer 1

Answer: (C)

$E=\frac{hc}{\lambda }$ or $E\propto \frac{1}{\lambda }\therefore \frac{E_2}{E_1}=\frac{{\lambda }_1}{{\lambda }_2}$
or $E_2=E_1\times \frac{{\lambda }_1}{{\lambda }_2}=1.23\times \frac{10,000}{5,000}=2.46eV$
Now, $hv-{\varphi }_0=\frac{1}{2}m{v}_{max}^2=e{V}_s$
$\Rightarrow {\varphi }_0=h{v}_2-e{V}_s=E_2-e{V}_s$
$=2.46-1.36=1.10eV$