MHT CET 2021 23rd September Evening Shift | Atoms and Nuclei Question 65 | Physics

Energy of electron in the second orbit of hydrogen atom is $$\mathrm{E}$$. The energy of electron '$$\mathrm{E}_3$$' in the third orbit of helium $$(\mathrm{He})$$ atom will be

$$\mathrm{E}_3=\frac{4 \mathrm{E}}{9}$$

$$\mathrm{E}_3=\frac{16 \mathrm{E}}{3}$$

$$\mathrm{E}_3=\frac{16 \mathrm{E}}{9}$$

$$\mathrm{E}_3=\frac{4 \mathrm{E}}{3}$$

MHT CET 2021 23th September Morning Shift

MCQ (Single Correct Answer)

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The shortest wavelength for Lyman series is 912 $$\mathop A\limits^o $$. The longest wavelength in Paschen series is

1216 $$\mathop A\limits^o $$

3646 $$\mathop A\limits^o $$

18760 $$\mathop A\limits^o $$

8208 $$\mathop A\limits^o $$

MHT CET 2021 23th September Morning Shift

MCQ (Single Correct Answer)

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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)

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

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

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

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

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