MHT CET 2024 9th May Evening Shift | Dual Nature of Radiation Question 17 | Physics

Using Einstein's photoelectric equation, the graph between kinetic energy of emitted photoelectrons and the frequency of incident radiation is shown correctly by graph

MHT CET 2024 9th May Evening Shift Physics - Dual Nature of Radiation Question 16 English

MHT CET 2024 9th May Evening Shift

MCQ (Single Correct Answer)

-0

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 15 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 )

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

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

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

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

PREVIOUS NEXT

MHT CET Subjects