Question

Find the wavelength of the incident light if the stopping potential is 0.6 V.

• A. 326 nm
• B. 454 nm
• C. 524 nm
• D. 232 nm

Answer 1

Answer: (B)

454 nm

According to Einstein's photoelectric equation
$e{V}_0=h\upsilon -{\varphi }_0=\frac{hc}{\lambda }-{\varphi }_0$ $(\because \upsilon =\frac{c}{\lambda })$

$e{V}_0=\frac{hc}{\lambda }-{\varphi }_0$

or $\lambda =\frac{hc}{(e{V}_0+{\varphi }_0)}=\frac{1242eVnm}{(0.6+2.14)eV}=454nm$