Question

Figure shows the graph of stopping potential versus the frequency of a photosensitive metal. The plank's constant and work function of the metal are:

• A. $\left(\frac{ V _{2}+ V _{1}}{v_{2}+v_{2}}\right) e ,\left(\frac{ V _{2} v_{1}+ V _{1} v_{2}}{v_{2}+v_{1}}\right) e$
• B. $\left(\frac{ V _{2}+ V _{1}}{v_{2}-v_{1}}\right) e ,\left(\frac{ V _{2} v_{1}+ V _{1} v_{2}}{v_{2}-v_{1}}\right) e$
• C. $\left(\frac{ V _{2}- V _{1}}{v_{2}+v_{1}}\right) e ,\left(\frac{ V _{2} v_{1}- V _{1} v_{2}}{v_{2}+v_{1}}\right) e$
• D. $\left(\frac{ V _{2}- V _{1}}{v_{2}-v_{1}}\right) e ,\left(\frac{ V _{2} v_{1}- V _{1} v_{2}}{v_{2}-v_{1}}\right) e$

Answer 1

Answer: (D)

Equation of straight line
$\frac{V-V_{1}}{v-v_{1}}=\frac{V_{2}-V_{1}}{v_{2}-v_{1}}$
$\Rightarrow V=\left(\frac{V_{2}-V_{1}}{v_{2}-v_{1}}\right) v-\left(\frac{V_{2} v_{1}-V_{1} v_{2}}{v_{2}-v_{1}}\right)$
But $V=\left(\frac{h}{e}\right) v-\frac{\phi_{0}}{e}$
$\Rightarrow h=\left(\frac{V_{2}-V_{1}}{v_{2}-v_{1}}\right) e$ $\&$
$\phi_{0}=\left(\frac{ V _{2} v_{1}- V _{1} v_{2}}{v_{2}-v_{1}}\right) e$