MHT CET 2021 23th September Morning Shift | Atoms and Nuclei Question 35 | Physics

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)

$$ \left(\frac{\mathrm{e}}{\mathrm{m}_{\mathrm{e}}}\right) \frac{\mathrm{nh}}{2 \pi} $$

$$ \mathrm{\left(\frac{e}{m_e}\right) \frac{n^2 h}{2 \pi}} $$

$$ \mathrm{\left(\frac{e}{2 m_e}\right) \frac{n^2 h}{2 \pi}} $$

$$ \left(\frac{\mathrm{e}}{2 \mathrm{~m}_{\mathrm{e}}}\right) \frac{\mathrm{nh}}{2 \pi} $$

MHT CET 2021 23th September Morning Shift

MCQ (Single Correct Answer)

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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)

$$\frac{2 \pi}{h}$$

$$\frac{\mathrm{nh}}{2 \pi}$$

$$\frac{\mathrm{h}}{\pi}$$

$$\frac{3 \mathrm{~h}}{2 \pi}$$

MHT CET 2021 22th September Evening Shift

MCQ (Single Correct Answer)

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

$$2: 1$$

$$1:-1$$

$$+1: 1$$

$$-1: 2$$

MHT CET 2021 22th September Evening Shift

MCQ (Single Correct Answer)

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

$$\mathrm{E} \propto \mathrm{L}^{-1}$$

$$\mathrm{E} \propto \mathrm{L}$$

$$\mathrm{E} \propto \mathrm{L}^{-2}$$

$$\mathrm{E} \propto \mathrm{L}^2$$

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