Question

The maximum velocities of the photoelectrons ejected are $v$ and $2v$ for the incident light of wavelength $400nm$ and $250nm$ on a metal surface respectively. The work function of the metal in terms of Planck's constant $h$ and velocity of light $c$ is

• A. $hc\times {10}^{6}J$
• B. $2hc\times {10}^{6}J$
• C. $1.5hc\times {10}^{6}J$
• D. $2.5hc\times {10}^{6}J$
• E. $3hc\times {10}^{6}J$

Answer 1

Answer: (B)

$2hc\times {10}^{6}J$

Key Concept According to Einstein's photoelectric equation, i.e.
$E=W_0+K_{max}$
where, $K_{max}=\frac{1}{2}m{v}_{max}^2=$ maximum kinetic energy of emitted electrons.
Given,
${\lambda }_1=400nm=400\times {10}^{-9}m$
${\lambda }_2=250nm=250\times {10}^{-9}m$
In the first condition,
$\frac{hc}{400\times {10}^{-9}}=\frac{1}{2}m{v}^2+\varphi ....\left(i\right)$
In the second condition,
$\frac{hc}{250\times {10}^{-19}}=\frac{1}{2}m(2v{)}^2+\varphi$
$\frac{hc}{250\times {10}^{-19}}=\frac{1}{2}m4v^2+\varphi ....\left(ii\right)$
From the Eq. (i), we get
$\frac{1}{2}m{v}^2=\frac{hc}{400\times {10}^{-9}}-\varphi .....\left(iii\right)$
Putting the value of Eq. (iii) in Eq. (ii), we get
$\frac{hc}{250\times {10}^{-9}}=4(\frac{hc}{400\times {10}^{-9}}-\varphi )+\varphi$
$\frac{hc}{250\times {10}^{-9}}=\frac{4hc}{400\times {10}^{-9}}-4\varphi +\varphi$
$\Rightarrow 3\varphi =\frac{4hc}{400\times {10}^{-9}}-\frac{hc}{250\times {10}^{-9}}$
$\Rightarrow 3\pi =\frac{hc}{100\times {10}^{-9}}-\frac{hc}{250\times {10}^{-9}}$
$\Rightarrow 3\varphi =hc\times {10}^9(\frac{1}{100}-\frac{1}{250})$
$\Rightarrow 3\varphi =hc\times {10}^9(0.01-0.004)$
$\therefore 3\varphi =6hc\times {10}^6$
Work function of the metal,
$\varphi =2hc\times {10}^{6}J$.