When a metal surface is illuminated by light | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = \Phi + K_{max}$,where $\Phi$ is the work function and $K_{max} = \frac{1}{2}

Vedclass · Accepted Answer

$\frac{hc}{3 \lambda_1 \lambda_2}(4 \lambda_2 - \lambda_1)$

Vedclass · Answer

(D) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = \Phi + K_{max}$,where $\Phi$ is the work function and $K_{max} = \frac{1}{2}mv^2$.For wavelength $\lambda_1$ with velocity $V$: $\frac{hc}{\lambda_1} = \Phi + \frac{1}{2}mV^2$  --- $(1)$For wavelength $\lambda_2$ with velocity $2V$: $\frac{hc}{\lambda_2} = \Phi + \frac{1}{2}m(2V)^2 = \Phi + 2mV^2$  --- $(2)$From $(1)$,$\frac{1}{2}mV^2 = \frac{hc}{\lambda_1} - \Phi$. Multiplying by $4$,we get $2mV^2 = \frac{4hc}{\lambda_1} - 4\Phi$.Substitute this into $(2)$: $\frac{hc}{\lambda_2} = \Phi + \frac{4hc}{\lambda_1} - 4\Phi$.$\frac{hc}{\lambda_2} = \frac{4hc}{\lambda_1} - 3\Phi$.$3\Phi = \frac{4hc}{\lambda_1} - \frac{hc}{\lambda_2} = hc \left( \frac{4\lambda_2 - \lambda_1}{\lambda_1 \lambda_2} \right)$.$\Phi = \frac{hc}{3 \lambda_1 \lambda_2} (4 \lambda_2 - \lambda_1)$.

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,$\lambda_1 > \lambda_2$).

Similar Questions

The ratio of work functions of two metals is $1:2$. If light of frequencies $f$ and $2f$ are incident on them respectively,what is the ratio of the maximum kinetic energy of the emitted photoelectrons?

The work function of a metallic surface is $5.01 \, eV$. The photo-electrons are emitted when light of wavelength $2000 \, \mathring{A}$ falls on it. The potential difference applied to stop the fastest photo-electrons is ............... $volt$ $[h = 4.14 \times 10^{-15} \, eV \cdot s]$

$A$ light whose frequency is equal to $6 \times 10^{14} \, Hz$ is incident on a metal whose work function is $2 \, eV$. $[h = 6.63 \times 10^{-34} \, Js, 1 \, eV = 1.6 \times 10^{-19} \, J]$. The maximum kinetic energy of the emitted electrons will be ............ $eV$. (in $.49$)

Photoelectrons are emitted with maximum velocity $v$ when light of frequency $3f$ is incident on a photosensitive material of work function $2hf$. If the frequency of the incident light is $4.25f$,the maximum velocity of the emitted photoelectrons is ($h$ = Planck's constant).

The kinetic energy of an emitted electron is $E$ when the light incident on the metal has wavelength $\lambda$. To double the kinetic energy,the incident light must have a wavelength of:

Vedclass Test Series

Exam Paper Generator

Online Exam Module