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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by $K_{\max} = \frac{hc}{\lambda} - \phi$,where $\ph

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$9 \lambda$

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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by $K_{\max} = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$ is the work function and $\lambda_0$ is the threshold wavelength.Since $K_{\max} = eV_0$,we have $eV_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$.For the first case: $8e = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \quad \dots (i)$For the second case: $2e = \frac{hc}{3\lambda} - \frac{hc}{\lambda_0} \quad \dots (ii)$Multiply equation $(ii)$ by $3$: $6e = \frac{hc}{\lambda} - \frac{3hc}{\lambda_0} \quad \dots (iii)$Subtracting $(iii)$ from $(i)$: $(8e - 6e) = (\frac{hc}{\lambda} - \frac{hc}{\lambda}) - (\frac{hc}{\lambda_0} - \frac{3hc}{\lambda_0})$$2e = \frac{2hc}{\lambda_0}$$e = \frac{hc}{\lambda_0} \implies \frac{hc}{\lambda_0} = e$Substitute this into equation $(i)$: $8e = \frac{hc}{\lambda} - e \implies 9e = \frac{hc}{\lambda} \implies \frac{hc}{\lambda} = 9e$Since $\frac{hc}{\lambda_0} = e$ and $\frac{hc}{\lambda} = 9e$,we get $\frac{hc}{\lambda_0} = \frac{1}{9} \frac{hc}{\lambda}$,which implies $\lambda_0 = 9\lambda$.

When a metal surface is illuminated by light of wavelength $\lambda$,the stopping potential is $8 \ V$. When the same surface is illuminated by light of wavelength $3 \lambda$,the stopping potential is $2 \ V$. The threshold wavelength for this surface is:

Similar Questions

Light rays consisting of photons with energy $1.8 \ eV$ are incident on a metal surface with a work function of $1.2 \ eV$. What is the stopping potential required to stop the emitted electrons in $eV$?

$A$ photosensitive metallic surface has a work function $h\nu_0$. If photons of energy $2h\nu_0$ fall on this surface,the electrons are emitted with a maximum velocity of $4 \times 10^6 \, m/s$. When the photon energy is increased to $5h\nu_0$,the maximum velocity of the photoelectrons will be:

What happens to the incident photon involved in the photoelectric effect experiment?

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