Question 250: Dual Nature of Matter and Radiation

Question

Graph shows variation of stopping potential with frequency of incident radiation on a metal plate. The value of Planck's constant is [ $\mathrm{e}=$ charge on photoelectron]

Accepted Answer

$$ e V_o=h 
u-\phi $$
From graph
At C,
$$ \mathrm{eV}_1=\mathrm{h} v_1-\phi \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$$
At D,
$$ e V_2=h v_2-\phi \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$$
Subtracting eq (ii) from (i),
$$ \begin{aligned} & e V_1-e V_2=h v_1-h v_2 \ & h=\frac{e\left(V_1-V_2ight)}{\left(v_1-v_2ight)} \end{aligned} $$

Answer

$\frac{e\left(V_2-V_1\right)}{v_1 v_2}$

Answer

$\frac{e V_1 V_2}{\left(v_2-v_1\right)}$

Answer

$\frac{e\left(V_2-V_1\right)}{\left(v_2-v_1\right)}$

Answer

$\frac{e\left(V_1 V_2\right)}{v_1 v_2}$

Dual Nature of Matter and Radiation

Graph shows variation of stopping potential with frequency of incident radiation on a metal plate. The value of Planck's constant is [ \mathrm{e}= charge on photoelectron]

Choose the correct answer: