Question

The maximum velocities of the photoelectrons ejected are $v$ and $2v$ for the incident light of wavelength $400\, nm$ and $250 \,nm$ 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. $ 2 hc \times 10^6 J$
• C. $ 1.5 hc \times 10^6 J$
• D. $2.5 hc \times 10^6 J$
• E. $3 hc \times 10^6 J$

Answer 1

Answer: (B)

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,
$\lambda_{1}=400\, nm =400 \times 10^{-9} m$
$\lambda_{2}=250\, nm =250 \times 10^{-9} m$
In the first condition,
$\frac{h c}{400 \times 10^{-9}}=\frac{1}{2} m v^{2}+\phi$ ...(i)
In the second condition,
$\frac{h c}{250 \times 10^{-19}}=\frac{1}{2} m(2 v)^{2}+\phi$
$\frac{h c}{250 \times 10^{-19}}=\frac{1}{2} m 4 v^{2}+\phi$ ...(ii)
From the Eq. (i), we get
$\frac{1}{2} m v^{2}=\frac{h c}{400 \times 10^{-9}}-\phi$ ...(iii)
Putting the value of Eq. (iii) in Eq. (ii), we get
$\frac{h c}{250 \times 10^{-19}}=4\left(\frac{h c}{400 \times 10^{-19}}-\phi\right)+\phi$
$\frac{h c}{250 \times 10^{-9}}=\frac{4 h c}{400 \times 10^{-9}}-4 \phi+\phi$
$\Rightarrow 3 \phi=\frac{4 h c}{400 \times 10^{-9}}-\frac{h c}{250 \times 10^{-9}}$
$\Rightarrow 3 \phi=\frac{h c}{100 \times 10^{-9}}-\frac{h c}{250 \times 10^{-9}}$
$\Rightarrow 3 \phi=h c \times 10^{9}\left(\frac{1}{100}-\frac{1}{250}\right)$
$\Rightarrow 3 \phi=h c \times 10^{9}(0.01-0.004)$
$\therefore 3 \phi=6 h c \times 10^{6}$
Work function of the metal,
$\phi=2 h c \times 10^{6}\, J$