(D) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K.E._{\max} = hF - \Phi$,where $\Phi = hF_0$ is the work fu

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K.E._{\max} = hF - \Phi$,where $\Phi = hF_0$ is the work function.Case $1$: When incident frequency $F = 2F_0$,$\frac{1}{2}mV_1^2 = h(2F_0) - hF_0 = hF_0$Case $2$: When incident frequency $F = 5F_0$,$\frac{1}{2}mV_2^2 = h(5F_0) - hF_0 = 4hF_0$Dividing the two equations:$\frac{V_1^2}{V_2^2} = \frac{hF_0}{4hF_0} = \frac{1}{4}$Taking the square root on both sides:$\frac{V_1}{V_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}$

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

Medium

The threshold frequency of a metal is $F_0$. When light of frequency $2F_0$ is incident on the metal plate,the maximum velocity of the photoelectrons is $V_1$. When the frequency of incident radiation is increased to $5F_0$,the maximum velocity of the photoelectrons emitted is $V_2$. The ratio of $V_1$ to $V_2$ is:

MHT CET 2024 All MHT CET

A
$1:8$
B
$1:16$
C
$1:4$
D
$1:2$

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Dual Nature of Radiation and matter

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Einstein's Photoelectric Equation and Energy Quantum of Radiation

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When a metal surface is illuminated by light of wavelengths $400\; nm$ and $250\; nm$,the maximum velocities of the photoelectrons ejected are $v$ and $2v$ respectively. The work function of the metal is ($h =$ Planck's constant,$c =$ velocity of light in air).

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$A$ and $B$ are two metals with threshold frequencies $1.8 \times 10^{14} \ Hz$ and $2.2 \times 10^{14} \ Hz$. Two identical photons of energy $0.825 \ eV$ each are incident on them. Then photoelectrons are emitted by (Take $h = 6.6 \times 10^{-34} \ J \cdot s$)

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Which one of the four graphs showing lines $P, Q, R$ and $S$ between maximum kinetic energy $(E)$ and intensity of incident radiation $(I)$ is correct?

When electromagnetic radiation is incident on a metallic surface,the maximum kinetic energy of the photoelectrons depends on .......

In a photoelectric experiment,a graph is drawn with stopping potential along $Y$-axis and the frequency of the incident light along $X$-axis. If the graph is a straight line which makes an angle $\theta$ with $Y$-axis,then $\tan \theta=$ ($h$-Planck's constant,$e$-charge of electron).

When a metal surface is illuminated by light of wavelengths $400\; nm$ and $250\; nm$,the maximum velocities of the photoelectrons ejected are $v$ and $2v$ respectively. The work function of the metal is ($h =$ Planck's constant,$c =$ velocity of light in air).

$A$ and $B$ are two metals with threshold frequencies $1.8 \times 10^{14} \ Hz$ and $2.2 \times 10^{14} \ Hz$. Two identical photons of energy $0.825 \ eV$ each are incident on them. Then photoelectrons are emitted by (Take $h = 6.6 \times 10^{-34} \ J \cdot s$)

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