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  • Maximum velocity of the photoelectrons emitted by a metal surface is <img alt="1.2 \times 10^6 \mathrm{~ms}^{-1}" src="https://entrancecorner.oncodecogs.com/gif.latex?1.2%20%5Ctimes%2010%5E6%20%5Cm

Maximum velocity of the photoelectrons emitted by a metal surface is 1.2 \times 10^6 \mathrm{~ms}^{-1}. Assuming the specific charge of the electron to be1.8 \times 10^{11} \mathrm{C} \mathrm{Kg}^{-1}, the value of the stopping potential in volt will be -

Option: 1

2


Option: 2

3


Option: 3

4


Option: 4

6


Answers (1)

best_answer

\begin{aligned} & \frac{1}{2} \mathrm{mv}^2=\mathrm{eV}_0 \\ & \mathrm{~V}_0=\frac{1}{2} \frac{\mathrm{m}}{\mathrm{e}} \mathrm{v}^2 \\ & \mathrm{~V}_0=\frac{1}{2} \times \frac{1}{1.8 \times 10^{11}} \times\left(1.2 \times 10^6\right)^2=4 \mathrm{Volt} \end{aligned}

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