(A) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K.E_{\max} = E - \phi$. For the first case,$E_1 = 3\phi$,so

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(A) According to Einstein's photoelectric equation,the maximum kinetic energy is given by $K.E_{\max} = E - \phi$. For the first case,$E_1 = 3\phi$,so $K.E_1 = 3\phi - \phi = 2\phi$. For the second case,$E_2 = 9\phi$,so $K.E_2 = 9\phi - \phi = 8\phi$. Since $K.E = \frac{1}{2}mv^2$,we have the ratio $\frac{K.E_1}{K.E_2} = \frac{v_1^2}{v_2^2}$. Substituting the values,$\frac{2\phi}{8\phi} = \frac{1}{4} = \frac{v_1^2}{v_2^2}$. Taking the square root,$\frac{v_1}{v_2} = \frac{1}{2}$,which implies $v_2 = 2v_1$. Given $v_1 = 6 \times 10^6 \ m/s$,then $v_2 = 2 \times 6 \times 10^6 = 12 \times 10^6 \ m/s$.

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

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$A$ photosensitive metallic surface has a work function $\phi$. If a photon of energy $3 \phi$ falls on the surface,the electron is emitted with a maximum velocity of $6 \times 10^6 \ m/s$. When the photon energy is increased to $9 \phi$,the maximum velocity of the photoelectrons will be:

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A
$12 \times 10^6 \ m/s$
B
$6 \times 10^6 \ m/s$
C
$3 \times 10^6 \ m/s$
D
$24 \times 10^6 \ m/s$

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Einstein's Photoelectric Equation and Energy Quantum of Radiation

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$A$ photosensitive metallic surface has a work function $\phi$. If a photon of energy $3 \phi$ falls on the surface,the electron is emitted with a maximum velocity of $6 \times 10^6 \ m/s$. When the photon energy is increased to $9 \phi$,the maximum velocity of the photoelectrons will be:

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An electromagnetic wave of wavelength $\lambda$ is incident on a photosensitive surface of negligible work function. If the photoelectron emitted from the surface has mass $m$ and de-Broglie wavelength $\lambda_{d}$,then

The study of photoelectric effect is useful in understanding

The de Broglie wavelength of the most energetic photoelectrons emitted from a photosensitive metal of work function $\phi$,when light of frequency $\nu$ is incident on it,is $\lambda$. Then $\nu =$ (where $h$ is Planck's constant and $m$ is the mass of the electron).

Assertion : Photosensitivity of a metal is high if its work function is small.Reason : Work function $= hf_0$ where $f_0$ is the threshold frequency.

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Assertion : Photosensitivity of a metal is high if its work function is small.
Reason : Work function $= hf_0$ where $f_0$ is the threshold frequency.