Question

Stopping potential for photoelectrons ejected from a surface when light of wavelength $500nm$ falls on it is $0.19V$. Now if light of wavelength $200nm$ falls on same surface stopping potential will be (in volts) \(take $hc=1250eVnm$) (Write up to two digits after the decimal point.)

Answer 1

$0.19eV=\frac{1250}{500}-\varphi ....\left(1\right)$
$VeV=\frac{1250}{200}-\varphi ....\left(2\right)$
By (2) - (1) $\Rightarrow (V-0.19)eV=1250(\frac{1}{200}-\frac{1}{500})eVV=0.19+1250\times \frac{300}{200\times 500}=3.94V$