Question

Radiation of wavelength, ${}^{\prime }{\lambda }^{\prime }$ is incident on a photocell. The fastest emitted electron has speed ‘u’. If the wavelength is changed to $\frac{3\lambda }{4}$, the speed of the fastest emitted electron will be

• A. $=u{\left(\frac{3}{4}\right)}^{\frac{1}{2}}$
• B. $>u{\left(\frac{4}{3}\right)}^{\frac{1}{2}}$
• C. $>u{\left(\frac{3}{4}\right)}^{\frac{1}{2}}$
• D. $=u{\left(\frac{4}{3}\right)}^{\frac{1}{2}}$

Answer 1

The correct option is B $>u{\left(\frac{4}{3}\right)}^{\frac{1}{2}}$

We can formulate the given situation as :

$E-\varphi =K_1$

$\frac{4}{3}E-\varphi =K_2$ (as new wavelength is $\frac{3}{4}$ times the initial wavelength)

Let the speed of fastest electron be $u$ and $u_1$ respectively.

On arranging the above equations, we get :
$u_1=\sqrt{\frac{4}{3}}\sqrt{\frac{E-\frac{3}{4}\varphi }{E-\varphi }}u$
which concludes that $u_1>u{\left(\frac{4}{3}\right)}^{\frac{1}{2}}$