(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = E - \Phi$,where $E$ is the energy of th

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = E - \Phi$,where $E$ is the energy of the incident photon and $\Phi$ is the work function.Case $1$: $E_1 = 2hv_0$ and $\Phi = hv_0$.$K_{max1} = 2hv_0 - hv_0 = hv_0 = \frac{1}{2}mv_1^2$,where $v_1 = 4 \times 10^6 \, m/s$.Case $2$: $E_2 = 5hv_0$ and $\Phi = hv_0$.$K_{max2} = 5hv_0 - hv_0 = 4hv_0 = \frac{1}{2}mv_2^2$.Dividing the two equations:$\frac{\frac{1}{2}mv_2^2}{\frac{1}{2}mv_1^2} = \frac{4hv_0}{hv_0} = 4$.$\frac{v_2^2}{v_1^2} = 4 \Rightarrow \frac{v_2}{v_1} = 2$.$v_2 = 2v_1 = 2 \times (4 \times 10^6 \, m/s) = 8 \times 10^6 \, m/s$.

Class 12 Physics Dual Nature of Radiation and matter Einstein's Photoelectric Equation and Energy Quantum of Radiation

Medium

The work function of a photosensitive metal surface is $hv_0$. If a photon of energy $2hv_0$ is incident on this surface,electrons are emitted with a maximum velocity of $4 \times 10^6 \, m/s$. What will be the maximum velocity of the photoelectrons when the energy of the incident photon is increased to $5hv_0$?

A
$2 \times 10^6 \, m/s$
B
$2 \times 10^7 \, m/s$
C
$8 \times 10^5 \, m/s$
D
$8 \times 10^6 \, m/s$

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Dual Nature of Radiation and matter

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Einstein's Photoelectric Equation and Energy Quantum of Radiation

The stopping potential as a function of frequency of incident radiation is plotted for two different photoelectric surfaces $A$ and $B$. The graph shows that the work function of $A$ is

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$A$ photoemissive substance is illuminated with a radiation of wavelength $\lambda_i$ so that it releases electrons with de-Broglie wavelength $\lambda_e$. The longest wavelength of radiation that can emit photoelectron is $\lambda_0$. The expression for the de-Broglie wavelength $\lambda_e$ is ($m = \text{mass of electron}$, $h = \text{Planck's constant}$, $c = \text{speed of light}$):

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$A$ photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $(\lambda/3)$. If the maximum kinetic energy of the emitted photoelectrons in the second case is $4$ times that in the first case, the work function of the surface of the material is ($h$ = Planck's constant, $c$ = speed of light).

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Ultraviolet light of wavelength $200 \ nm$ is incident on a freshly polished surface of iron. The work function of the surface is $4.71 \ eV$. What will be the stopping potential in $eV$? $(h = 6.626 \times 10^{-34} \ Js, 1 \ eV = 1.6 \times 10^{-19} \ J, c = 3 \times 10^8 \ m/s)$

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According to Einstein's photoelectric equation,the graph of the kinetic energy of photoelectrons emitted from a metal versus the frequency of incident radiation is a straight line whose slope:

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The stopping potential as a function of frequency of incident radiation is plotted for two different photoelectric surfaces $A$ and $B$. The graph shows that the work function of $A$ is

Ultraviolet light of wavelength $200 \ nm$ is incident on a freshly polished surface of iron. The work function of the surface is $4.71 \ eV$. What will be the stopping potential in $eV$? $(h = 6.626 \times 10^{-34} \ Js, 1 \ eV = 1.6 \times 10^{-19} \ J, c = 3 \times 10^8 \ m/s)$

According to Einstein's photoelectric equation,the graph of the kinetic energy of photoelectrons emitted from a metal versus the frequency of incident radiation is a straight line whose slope:

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