Question

In a photo emissive cell, with exciting wavelength $\lambda$, the maximum kinetic energy of electron is $K$. If the exciting wavelength is changed to$\frac{3\lambda }{4}$, the kinetic energy of the fastest emitted electron will be

• A. $3K/4$
• B. $4K/3$
• C. less than $4K/3$
• D. greater than $4K/3$

Answer 1

The correct option is D greater than $4K/3$

From Einstein's photo - electric equation,
$\frac{hc}{\lambda }=W+K\Rightarrow K=\frac{hc}{\lambda }(1-\frac{W\lambda }{hc})\text{and}\frac{hc}{\left(\frac{3\lambda }{4}\right)}=W+K_1\Rightarrow K_1=\frac{4hc}{3\lambda }(1-\frac{3W\lambda }{4hc})$

Dividing,
$\frac{K_1}{K}=\frac{4}{3}\times f\text{where}f=\frac{1-\frac{3W\lambda }{4hc}}{1-\frac{W\lambda }{hc}}>1\Rightarrow K_1=\left(\frac{4K}{3}\right)f$
Since $f>1\Rightarrow K_1>\frac{4K}{3}$