In a photoelectric effect experiment,the cath | vedclass

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

Question

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

Vedclass · Accepted Answer

$1.02$

Vedclass · Answer

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.For $\lambda_1 = 600 \ nm$,$K_1 = \frac{hc}{\lambda_1} - \phi$  --- $(i)$For $\lambda_2 = 400 \ nm$,$K_2 = 2K_1 = \frac{hc}{\lambda_2} - \phi$ --- (ii)From $(i)$,$K_1 = \frac{hc}{\lambda_1} - \phi$. Substituting this into (ii):$2 \left( \frac{hc}{\lambda_1} - \phi ight) = \frac{hc}{\lambda_2} - \phi$$\frac{2hc}{\lambda_1} - 2\phi = \frac{hc}{\lambda_2} - \phi$$\phi = hc \left( \frac{2}{\lambda_1} - \frac{1}{\lambda_2} ight)$Substituting the values: $h = 6.63 	imes 10^{-34} \ J-s$,$c = 3 	imes 10^8 \ m/s$,$\lambda_1 = 600 	imes 10^{-9} \ m$,$\lambda_2 = 400 	imes 10^{-9} \ m$:$\phi = (6.63 	imes 10^{-34} 	imes 3 	imes 10^8) \left( \frac{2}{600 	imes 10^{-9}} - \frac{1}{400 	imes 10^{-9}} ight)$$\phi = (19.89 	imes 10^{-26}) 	imes 10^9 \left( \frac{1}{300} - \frac{1}{400} ight)$$\phi = 19.89 	imes 10^{-17} \left( \frac{4-3}{1200} ight) = \frac{19.89 	imes 10^{-17}}{1200} \ J$$\phi = \frac{19.89 	imes 10^{-17}}{1200 	imes 1.6 	imes 10^{-19}} \ eV \approx 1.036 \ eV \approx 1.02 \ eV$.

Similar Questions

The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength $491\, nm$ is $0.710\, V$. When the incident wavelength is changed to a new value,the stopping potential is $1.43\, V$. The new wavelength is ....... $nm.$

The maximum kinetic energies of photoelectrons emitted are $K_1$ and $K_2$ when lights of wavelengths $\lambda_1$ and $\lambda_2$ are incident on a metallic surface. If $\lambda_1 = 3 \lambda_2$,then:

The photoelectric work function for a metal surface is $4.125 \ eV$. The cut-off wavelength for this surface is .......... $\mathring{A}$.

When a point source of light is at a distance of $1 \ m$ from a photocell,the cut-off voltage is found to be $V$. If the same source is placed at $2 \ m$ distance from the photocell,the cut-off voltage will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module