Question

Light of wavelength $2000\stackrel{\circ }{\text{A}}$ is incident on the cathode of a photocell. The current in the photocell is reduced to zero by stopping potential of $2\text{V}$. Find the threshold wavelength of the material of the cathode.
$\text{Given}:\text{Planck's constamt}\left(h\right)=6.63\times {10}^{-34}{\text{J s}}^{},c=3\times {10}^8{\text{m s}}^{-1},e=1.6\times {10}^{-19}\text{C}$

• A. $294.9\stackrel{\circ }{\text{A}}$
• B. $2494\stackrel{\circ }{\text{A}}$
• C. $244.9\stackrel{\circ }{\text{A}}$
• D. $2949\stackrel{\circ }{\text{A}}$

Answer 1

The correct option is D $2949\stackrel{\circ }{\text{A}}$

By Einstein's photoelectric equation,
$K.E_{max}=hv-\Phi =\frac{hc}{\lambda }-\Phi$
But $K.E_{max}=e{V}_s$
$\Rightarrow e{V}_s=\frac{hc}{\lambda }-\Phi$

$\Rightarrow 2\times 1.6\times {10}^{-19}=\frac{(6.63\times {10}^{-34})(3\times {10}^8)}{(2000\times {10}^{-10})}-\Phi$

$\Rightarrow 3.2\times {10}^{-19}=9.945\times {10}^{-19}-\Phi$
​​​​​​​$\Rightarrow \Phi =6.745\times {10}^{-19}\text{J}$

Since $\Phi =h{v}_o=\frac{hc}{{\lambda }_o}$

$\therefore {\lambda }_o=\frac{(6.63\times {10}^{-34})(3\times {10}^8)}{(6.745\times {10}^{-19})}$

$\Rightarrow {\lambda }_o=2.949\times {10}^{-7}\text{m}=2949\stackrel{\circ }{\text{A}}$