MHT CET 2025 26th April Evening Shift | Dual Nature of Radiation Question 1 | Physics

Let $E_c$ and $E_p$ represents kinetic energy of electron and photon respectively. If de-Broglie wavelength of a photon is twice the de-Broglie wavelength of an electron then $E_p / E_c$ is (speed of electron $=\mathrm{C} / 100$ where C is the velocity of light)

$10^2$

$10^3$

$10^4$

MHT CET 2025 26th April Evening Shift

MCQ (Single Correct Answer)

-0

The graph shows the variation of photocurrent with anode potential for four different radiations. Let $\mathrm{I}_{\mathrm{a}}, \mathrm{I}_{\mathrm{b}}, \mathrm{I}_{\mathrm{c}}$ and $\mathrm{I}_{\mathrm{d}}$ are intensities and $\mathrm{f}_{\mathrm{a}}, \mathrm{f}_{\mathrm{b}}, \mathrm{f}_{\mathrm{c}}$ and $\mathrm{f}_{\mathrm{d}}$ be the frequencies for the curves $\mathrm{a}, \mathrm{b}, \mathrm{c}$ and d respectively, then

$\mathrm{f}_{\mathrm{b}}>\mathrm{f}_{\mathrm{a}}, \mathrm{f}_{\mathrm{b}}=\mathrm{f}_{\mathrm{c}}, \mathrm{I}_{\mathrm{c}}=\mathrm{I}_{\mathrm{d}}$

$\mathrm{f}_{\mathrm{b}}=\mathrm{f}_{\mathrm{a}}, \mathrm{f}_{\mathrm{b}}>\mathrm{f}_{\mathrm{c}}, \mathrm{I}_{\mathrm{c}}>\mathrm{I}_{\mathrm{d}}$

$\mathrm{f}_{\mathrm{b}}<\mathrm{f}_{\mathrm{a}}, \mathrm{f}_{\mathrm{b}}<\mathrm{f}_{\mathrm{c}}, \mathrm{I}_{\mathrm{c}}<\mathrm{I}_{\mathrm{d}}$

$\mathrm{f}_{\mathrm{b}}<\mathrm{f}_{\mathrm{a}}, \mathrm{f}_{\mathrm{b}}>\mathrm{f}_{\mathrm{c}}, \mathrm{I}_{\mathrm{c}}=\mathrm{I}_{\mathrm{d}}$

MHT CET 2025 26th April Morning Shift

MCQ (Single Correct Answer)

-0

Light of incident frequency 3 times the threshold frequency is incident on a photosensitive material. If the incident frequency is made $\left(\frac{1}{4}\right)^{\text {th }}$ and intensity is tripled then the photoelectric current will

increase.

decrease.

be $\left(\frac{1}{3}\right)^{\text {rd }}$

be zero.

MHT CET 2025 26th April Morning Shift

MCQ (Single Correct Answer)

-0

The wavelength ' $\lambda$ ' of a photon and the deBroglie wavelength of an electron have same value. the ratio of kinetic energy of the electron to the energy of a photon is

( $\mathrm{m}=$ mass of electron, $\mathrm{c}=$ velocity of light, $\mathrm{h}=$ Planck's constant)

$\frac{2 \lambda \mathrm{mc}}{\mathrm{h}}$

$\frac{\lambda \mathrm{mc}}{\mathrm{h}}$

$\frac{\mathrm{h}}{2 \lambda \mathrm{mc}}$

$\frac{\mathrm{h}}{\lambda \mathrm{mc}}$

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