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When a light of photons of energy 4.2 \mathrm{~eV}is incident on a metallic sphere of radius 10 \mathrm{~cm}and work function 2.4 \mathrm{~eV}, photoelectrons are emitted. The number of photoelectrons liberated before the emission is stopped, is 

\left(\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\right. \text{ and } \left.\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-2}\right)

 

Option: 1

6.25 \times 10^8


Option: 2

1.25 \times 10^9


Option: 3

1.25 \times 10^8


Option: 4

6.25 \times 10^{18}


Answers (1)

best_answer

Here,
Radius of the sphere, \mathrm{r}=10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}
Work function, \phi_0=2.4 \mathrm{eV}
Energy of a photon, \mathrm{E}=\mathrm{h} v=4.2 \mathrm{eV}
According to Einstein's photoelectric equation

\begin{aligned} & \mathrm{hv}=\phi_0+\mathrm{eV}_5 \\ & 4.2 \mathrm{eV}=2.4 \mathrm{eV}+\mathrm{eV}_5 \\ & \mathrm{eV}_{\mathrm{s}}=4.2 \mathrm{eV}-2.4 \mathrm{eV}=1.8 \mathrm{eV} \\ & \therefore \mathrm{V}_{\mathrm{s}}=1.8 \mathrm{~V} \end{aligned}

The sphere will stop emitting photoelectrons, when the potential on its surface becomes 1.8 \mathrm{~V}.

Let \mathrm{N}be the number of photoelectrons emitted from the sphere.

Then, Charge on the sphere, Q=\mathrm{Ne}

\begin{aligned} & \mathrm{V}_{\mathrm{s}}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Q}}{\mathrm{r}}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{Ne}}{\mathrm{T}} \\ & \mathrm{N}=\frac{\mathrm{V}_5 \times \mathrm{I}}{\frac{1}{4 \pi \varepsilon_0} \times \mathrm{e}}=\frac{1.8 \times \mathrm{r}}{\frac{1}{4 \pi \varepsilon_0} \times \mathrm{e}}=\frac{1.8 \times 10 \times 10^{-2}}{9 \times 10^9 \times 1.6 \times 10^{-19}}=\frac{18}{16} \times \frac{1}{9} \times 10^9=1.25 \times 10^8 \end{aligned}

Posted by

Pankaj Sanodiya

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