1
MHT CET 2021 23th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

In the Bohr model, an electron moves in a circular orbit around the nucleus. Considering an orbiting electron to be a circular current loop, the magnetic moment of the hydrogen atom, when the electron is in nth excited state, is

(e = electronic charge, m$$_e$$ = mass of the electron, h = Planck's constant)

A
$$\left(\frac{\mathrm{e}}{\mathrm{m}_{\mathrm{e}}}\right) \frac{\mathrm{nh}}{2 \pi}$$
B
$$\mathrm{\left(\frac{e}{m_e}\right) \frac{n^2 h}{2 \pi}}$$
C
$$\mathrm{\left(\frac{e}{2 m_e}\right) \frac{n^2 h}{2 \pi}}$$
D
$$\left(\frac{\mathrm{e}}{2 \mathrm{~m}_{\mathrm{e}}}\right) \frac{\mathrm{nh}}{2 \pi}$$
2
MHT CET 2021 23th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

The energy of an electron in the excited hydrogen atom is $$-3.4 \mathrm{~eV}$$. Then according to Bohr's theory, the angular momentum of the electron in that excited state is ($$\mathrm{h}=$$ Plank's constant)

A
$$\frac{2 \pi}{h}$$
B
$$\frac{\mathrm{nh}}{2 \pi}$$
C
$$\frac{\mathrm{h}}{\pi}$$
D
$$\frac{3 \mathrm{~h}}{2 \pi}$$
3
MHT CET 2021 22th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

In $$n^{\text {th }}$$ Bohr orbit, the ratio of the kinetic energy of an electron to the total energy of it, is

A
$$2: 1$$
B
$$1:-1$$
C
$$+1: 1$$
D
$$-1: 2$$
4
MHT CET 2021 22th September Evening Shift
MCQ (Single Correct Answer)
+1
-0

If '$$E$$' and '$$L$$' denote the magnitude of total energy and angular momentum of revolving electron in $$\mathrm{n}^{\text {th }}$$ Bohr orbit, then

A
$$\mathrm{E} \propto \mathrm{L}^{-1}$$
B
$$\mathrm{E} \propto \mathrm{L}$$
C
$$\mathrm{E} \propto \mathrm{L}^{-2}$$
D
$$\mathrm{E} \propto \mathrm{L}^2$$
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