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Given that a photon of light of wavelength 10,000^{\circ}  has an energy equal to 1.23 \mathrm{eV}. When light of wavelength 5000 \AA  and intensity \mathrm{I}_0 falls on a photoelectric cell, the surface current is 0.40 \times 10^{-6} \mathrm{~A} and the stopping potential is 1.36 \mathrm{~V}, then the work function is:
 

Option: 1

0.43 \mathrm{eV}


Option: 2

0.55 \mathrm{eV}


Option: 3

1.10 \mathrm{eV}


Option: 4

1.53 \mathrm{eV}


Answers (1)

best_answer

The energy of a photon E=\frac{h c}{\lambda}
\begin{aligned} & \text{or } \mathrm{E} \propto \frac{1}{\lambda} \quad \quad \quad \therefore \frac{E_2}{E_1}=\frac{\lambda_1}{\lambda_2} \\ & \text{or } \mathrm{E}_2=\mathrm{E}_1 \times \frac{\lambda_1}{\lambda_2}=1.23 \times \frac{10000}{5000}=2.46 \mathrm{eV} \end{aligned}

According to Einstein's photoelectric equation
\begin{aligned} & \mathrm{h\nu}-\phi_0=\frac{1}{2} \mathrm{mv}_{\max }^2=\mathrm{eV}_{\mathrm{s}} \\ & \text { or } \phi_0=\mathrm{h\nu}_2-\mathrm{eV}_{\mathrm{s}}=\mathrm{E}_2-\mathrm{eV}_{\mathrm{s}}=2.46-1.36=1.10 \mathrm{eV} \end{aligned}

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shivangi.shekhar

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