MHT CET 2024 9th May Evening Shift | Dual Nature of Radiation Question 44 | Physics

The figure shows the variation of photocurrent with anode potential for four different radiations. Let $f_a, f_b, f_c$ and $f_d$ be the frequencies for the curves $a, b, c$ and $d$ respectively

MHT CET 2024 9th May Evening Shift Physics - Dual Nature of Radiation Question 43 English

$\mathrm{f_a>f_b>f_c>f_d}$

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

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

$\mathrm{f_a=f_b>f_c>f_d}$

MHT CET 2024 9th May Morning Shift

MCQ (Single Correct Answer)

-0

The gyromagnetic ratio and Bohr magneton are given respectively by [Given $\rightarrow \mathrm{e}=$ charge on electron, $\mathrm{m}=$ mass of electron, $\mathrm{h}=$ Planck's constant]

$\frac{\mathrm{e}}{2 \mathrm{~m}}, \frac{\mathrm{eh}}{4 \pi \mathrm{~m}}$

$\frac{\mathrm{eh}}{4 \pi \mathrm{~m}}, \frac{\mathrm{e}}{2 \mathrm{~m}}$

$\frac{2 \mathrm{~m}}{\mathrm{e}}, \frac{4 \pi \mathrm{~m}}{\mathrm{eh}}$

$\frac{4 \pi \mathrm{~m}}{\mathrm{eh}}, \frac{2 \mathrm{~m}}{\mathrm{e}}$

MHT CET 2024 9th May Morning Shift

MCQ (Single Correct Answer)

-0

Two identical photocathodes receive light of frequencies ' $\mathrm{n}_1$ ' and ' $\mathrm{n}_2$ '. If the velocities of the emitted photoelectrons of mass ' $m$ ' are ' $\mathrm{V}_1$ ' and ' V , respectively, then ( $\mathrm{h}=$ Planck's constant )

$\mathrm{V}_1+\mathrm{V}_2=\left[\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{n}_1+\mathrm{n}_2\right)\right]^{1 / 2}$

$\mathrm{V}_1-\mathrm{V}_2=\left[\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{n}_1-\mathrm{n}_2\right)\right]^{1 / 2}$

$\mathrm{V}_1^2+\mathrm{V}_2^2=\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{n}_1+\mathrm{n}_2\right)$

$\mathrm{V}_1^2-\mathrm{V}_2^2=\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{n}_1-\mathrm{n}_2\right)$

MHT CET 2024 4th May Evening Shift

MCQ (Single Correct Answer)

-0

The kinetic energy of an electron is increased by 2 times, then the de-Broglie wavelength associated with it changes by a factor.

$\frac{1}{3}$

$\frac{1}{\sqrt{3}}$

$\sqrt{3}$

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