Question

The kinetic energy of the photoelectrons increases by $0.52eV$ when the wavelength of incident light is changed from $500nm$ to another wavelength which is approximately

• A. 700 nm
• B. 400 nm
• C. 1250 nm
• D. 1000 nm

Answer 1

Answer: (B)

Here, change in kinetic energy, $\Delta K=0.52eV,\lambda =500nm,{\lambda }_2=?$
We know , $K_1=\frac{hc}{{\lambda }_1}=\varphi$
$K_2=\frac{hc}{{\lambda }_2}-\varphi$
$\therefore K_1=K_2=hc\left(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}\right)$
or, $-\Delta K=hc\left(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}\right)$
or, $-0.52eV=\left(1242eVnm\right)\left(\frac{1}{500nm}-\frac{1}{{\lambda }_2}\right)$
or, $\frac{-0.52}{1242}=\frac{1}{500}-\frac{1}{{\lambda }_2}$
or,$\frac{1}{{\lambda }_2}=\frac{1}{500}+\frac{0.52}{1242}$
or, ${\lambda }_2=413nm=400nm$