Photoelectric effect experiments are performe | vedclass

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

Question

(A) Given work functions are $\phi_p=2.0 \ eV, \phi_q=2.5 \ eV$ and $\phi_r=3.0 \ eV$.Using the relation $\lambda_0 = \frac{hc}{\phi}$,where $hc = 124

Vedclass · Accepted Answer

Vedclass · Answer

(A) Given work functions are $\phi_p=2.0 \ eV, \phi_q=2.5 \ eV$ and $\phi_r=3.0 \ eV$.Using the relation $\lambda_0 = \frac{hc}{\phi}$,where $hc = 1240 \ eV \ nm$,we calculate the threshold wavelengths:$\lambda_p = \frac{1240}{2.0} = 620 \ nm$$\lambda_q = \frac{1240}{2.5} = 496 \ nm$$\lambda_r = \frac{1240}{3.0} \approx 413.3 \ nm$The incident light contains wavelengths $\lambda_1 = 550 \ nm, \lambda_2 = 450 \ nm, \lambda_3 = 350 \ nm$ with equal intensities.For photoelectric emission,we must have $\lambda \le \lambda_0$:- For plate $p$ $(\lambda_p = 620 \ nm)$: All three wavelengths $(550, 450, 350 \ nm)$ cause emission. Thus,the saturation current $I_p$ is proportional to the sum of intensities of all three wavelengths.- For plate $q$ $(\lambda_q = 496 \ nm)$: Only $\lambda_2 = 450 \ nm$ and $\lambda_3 = 350 \ nm$ cause emission. Thus,$I_q$ is proportional to the sum of intensities of two wavelengths.- For plate $r$ $(\lambda_r = 413.3 \ nm)$: Only $\lambda_3 = 350 \ nm$ causes emission. Thus,$I_r$ is proportional to the intensity of one wavelength.Since intensities are equal,$I_p > I_q > I_r$. The saturation current is highest for $p$ and lowest for $r$. The correct graph is $A$.

Similar Questions

The kinetic energy of emitted cathode rays is dependent on:

Which of the following effects support the quantum nature of $EM$ radiation? $(1)$ Photoelectric effect $(2)$ Compton effect $(3)$ Doppler effect $(4)$ Field effect

Consider the following statements: $(1)$ Both wave and particle nature are required for a complete understanding of light. $(2)$ Light cannot exhibit both wave and particle nature simultaneously in a single experiment.

For a body moving with relativistic speed,if the velocity is doubled,then

An electron is moving through a field. It is moving $(i)$ opposite to an electric field and $(ii)$ perpendicular to a magnetic field as shown. For each situation,determine the de-Broglie wavelength of the electron.

Vedclass Test Series

Exam Paper Generator

Online Exam Module