Question

The stopping potential $\left(V_0\right)$ versus frequency ($\nu$) plot of a substance is shown in figure, the threshold wavelength is :

• A. $5\times {10}^{14}\mathring{A}$
• B. $6000\mathring{A}$
• C. $5000\mathring{A}$
• D. Cannot be estimated from given data

Answer 1

The correct option is B $6000\mathring{A}$

From the graph, when $V_0=0$, the corresponding frequency is the threshold frequency. i.e ${\nu }_0=5\times {10}^{14}$ Hz

so, threshold wavelength $,{\lambda }_0=\frac{c}{{\nu }_0}=\frac{3\times {10}^8}{5\times {10}^{14}}=6\times {10}^{-7}m=6000\mathring{A}$