Question

Photons having energy $1eV$ and $2.5eV$ successively incident on a metal, having work function is $0.5eV$. The ratio of maximum speed of emitted electrons is

• A. $2:1$
• B. $1:2$
• C. $3:1$
• D. $1:3$

Answer 1

Answer: (B)

$1:2$

Key Concept According to Einstein photoelectric equation, we have
$hv=\varphi +\frac{1}{2}m{v}_{max}^2$
where, $\varphi$ is work function and $\frac{1}{2}m{v}_{max}^2$ is kinetic energy of a particle having velocity $v$, $h$ is Planck's constant and $v$ is frequency of light radiation.
The Einstein equation is given as
$E-\varphi =\frac{1}{2}m{v}_{max}^2$
$\frac{hc}{\lambda }-\varphi =\frac{1}{2}m{v}_{max}^2$
In case of photon $1$ having energy $1eV$
$(1-0.5)eV=\frac{1}{2}m{v}_{max}^2$
$\Rightarrow 0.5eV=\frac{1}{2}m{v}_{max}^2....\left(i\right)$
Similarly, in another case, if photon $2$ having energy $25eV$
$(2.5-0.5)eV=\frac{1}{2}m{v}_{max}^2$
$2eV=\frac{1}{2}m{v}_{max}^{\prime 2}....\left(ii\right)$
Now, on dividing Eq. (ii) by Eq. (i), we get
$\frac{2eV}{\frac{1}{2}eV}=\frac{\frac{1}{2}m{v}_{max}^{\prime 2}}{\frac{1}{2}m{v}_{max}^2}$
$\Rightarrow \sqrt{4}=\frac{v_{max}^{\prime }}{v_{max}}$ or $\frac{v_{max}^{\prime 2}}{v_{max}^2}=2$
Hence, required ratio
$\frac{v_{max}}{v_{max}^{\prime }}=\frac{1}{2}=1:2$.