Two identical metal plates show the photoelec | vedclass

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

Question

(B) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = W_0 + K_{\max}$.For plate $A$: $\frac{hc}{\lambda_A} = W_0 + K_A$ ... $(i)$F

Vedclass · Accepted Answer

${K_A} < {K_B}/2$

Vedclass · Answer

(B) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = W_0 + K_{\max}$.For plate $A$: $\frac{hc}{\lambda_A} = W_0 + K_A$ ... $(i)$For plate $B$: $\frac{hc}{\lambda_B} = W_0 + K_B$ ... $(ii)$Given ${\lambda _A} = 2{\lambda _B}$,substitute this into $(i)$:$\frac{hc}{2\lambda_B} = W_0 + K_A$ ... $(iii)$From $(ii)$,we have $W_0 = \frac{hc}{\lambda_B} - K_B$. Substitute this into $(iii)$:$\frac{hc}{2\lambda_B} = (\frac{hc}{\lambda_B} - K_B) + K_A$Rearranging the terms:$K_A = K_B - \frac{hc}{2\lambda_B}$Since $\frac{hc}{\lambda_B} = W_0 + K_B$,then $\frac{hc}{2\lambda_B} = \frac{W_0 + K_B}{2}$.Substituting this back:$K_A = K_B - \frac{W_0 + K_B}{2} = \frac{2K_B - W_0 - K_B}{2} = \frac{K_B - W_0}{2} = \frac{K_B}{2} - \frac{W_0}{2}$.Since $W_0 > 0$,it follows that $K_A < \frac{K_B}{2}$.

Two identical metal plates show the photoelectric effect. Light of wavelength ${\lambda _A}$ falls on plate $A$ and light of wavelength ${\lambda _B}$ falls on plate $B$,where ${\lambda _A} = 2{\lambda _B}$. The maximum kinetic energy is:

Similar Questions

Two photons, each having an energy of $2.5 \ eV$, are incident on a metal surface with a work function of $4.5 \ eV$. Then . . . . . . .

The threshold wavelength for photoelectric emission in tungsten is $400 \ nm$. The wavelength of light that must be used in order to eject electrons with a maximum kinetic energy of $0.9 \ eV$ is .............. $nm$.

Photoelectrons are ejected from a metal when light of frequency $v$ falls on it. Pick out the wrong statement from the following:

In an experiment aimed to disprove Einstein's photoelectric equation,how did Millikan prove it?

When a photon of wavelength $3000 \ \mathring{A}$ is incident on a metal surface,the energy of the emitted electron is $0.5 \ eV$. If a photon of wavelength $2000 \ \mathring{A}$ is incident on the same metal surface,the energy of the emitted electron will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module