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(C) As photoelectrons are emitted,the sphere becomes positively charged. The emission stops when the potential $V$ of the sphere is such that the maxi

Vedclass · Accepted Answer

$\frac{4\pi \epsilon_0 ahc}{e^2} \left[ \frac{1}{\lambda} - \frac{1}{\lambda_0} \right]$

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(C) As photoelectrons are emitted,the sphere becomes positively charged. The emission stops when the potential $V$ of the sphere is such that the maximum kinetic energy of the emitted photoelectrons is equal to the work done against the electrostatic potential.According to Einstein's photoelectric equation: $K_{max} = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$.The emission stops when $eV = K_{max}$,where $V$ is the potential of the sphere.Thus,$V = \frac{hc}{e} \left[ \frac{1}{\lambda} - \frac{1}{\lambda_0} \right]$.The charge $Q$ on the sphere of radius $a$ is given by $Q = CV = (4\pi \epsilon_0 a) V$.Substituting $V$: $Q = 4\pi \epsilon_0 a \cdot \frac{hc}{e} \left[ \frac{1}{\lambda} - \frac{1}{\lambda_0} \right]$.The number of photoelectrons emitted $n$ is given by $n = \frac{Q}{e} = \frac{4\pi \epsilon_0 ahc}{e^2} \left[ \frac{1}{\lambda} - \frac{1}{\lambda_0} \right]$.

Monochromatic light of wavelength $\lambda$ is incident on an isolated metal sphere of radius $a$. The threshold wavelength is $\lambda_0$,which is greater than $\lambda$. How many photoelectrons will be emitted before the emission of photoelectrons stops?

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