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

Vedclass · Accepted Answer

$4$

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(A) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K_{max} = eV_s = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function and $\lambda_t$ is the threshold wavelength such that $\phi = \frac{hc}{\lambda_t}$.For the first case: $e(3V_0) = \frac{hc}{\lambda} - \phi$  --- $(i)$For the second case: $eV_0 = \frac{hc}{2\lambda} - \phi$  --- (ii)Multiply equation (ii) by $3$: $3eV_0 = \frac{3hc}{2\lambda} - 3\phi$  --- (iii)Equating $(i)$ and (iii): $\frac{hc}{\lambda} - \phi = \frac{3hc}{2\lambda} - 3\phi$$2\phi = \frac{3hc}{2\lambda} - \frac{hc}{\lambda} = \frac{hc}{2\lambda}$$\phi = \frac{hc}{4\lambda}$Since $\phi = \frac{hc}{\lambda_t}$,we have $\frac{hc}{\lambda_t} = \frac{hc}{4\lambda}$,which implies $\lambda_t = 4\lambda$.Thus,the value of $n$ is $4$.

Similar Questions

The maximum velocity of the photoelectrons emitted by a metal surface is $9 \times 10^5 \ m/s$. The value of the ratio of charge $(e)$ to mass $(m)$ of the photoelectron is $1.8 \times 10^{11} \ C/kg$. The value of the stopping potential in volts is:

The photoelectric threshold wavelength of silver is $3250 \times 10^{-10} \, m$. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536 \times 10^{-10} \, m$ is (Given $h = 4.14 \times 10^{-15} \, eV \cdot s$ and $c = 3 \times 10^8 \, m/s$):

When a metal surface is illuminated by light of wavelength $\lambda_1$ and $\lambda_2$,the maximum velocities of photoelectrons ejected are $V$ and $2V$ respectively. The work function of the metal is ($h=$ Planck's constant,$c=$ velocity of light,$\lambda_1 > \lambda_2$).

$4 \ eV$ is the energy of the incident photon and the work function is $2 \ eV$. What is the stopping potential in $V$?

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