Question

If a radiation of wavelength $470\text{nm}$ falls on the surface of potassium metal, eletcrons are emitted with a maximum velocity of $6.4\times {10}^4{\text{m s}}^{-1}$.
Stopping potential of the given circuit is:

• A. 1.21 V
• B. 0.131 V
• C. 1.16 V
• D. 0.116 V

Answer 1

The correct option is D 0.116 V

Given: $\lambda =470\text{nm}$
$v=6.4\times {10}^4{\text{m s}}^{-1}$

Stopping potential is the required potential to cause zero current
Therefore,
$K.E.=\frac{1}{2}m{\text{v}}^2=\frac{1}{2}\times 9.1\times {10}^{-31}\times (6.4\times {10}^8{)}^2$

$=1.86\times {10}^{-20}\text{J}$

Again we know,
$K{E}_{max}=e{V}_s$
where, $V_s$ is the stopping potential
$1.86\times {10}^{-20}=1.6\times {10}^{-19}\times V_s$
$\Rightarrow V_s=\frac{1.86\times {10}^{-20}}{1.6\times {10}^{-19}}=0.116V$