MHT CET 2022 11th August Evening Shift | Atoms and Nuclei Question 27 | Physics

Using Bohr's model, the orbital period of electron in hydrogen atom in the $$\mathrm{n}^{\text {th }}$$ orbit is $$\left(\varepsilon_0=\right.$$ permittivity of vacuum, $$\mathrm{h}=$$ Planck's constant, $$\mathrm{m}=$$ mass of electron, $$\mathrm{e}=$$ electronic charge)

$$\frac{4 \varepsilon_0 \mathrm{nh}^3}{\mathrm{me}^2}$$

$$\frac{4 \varepsilon_0 \mathrm{n}^2 \mathrm{~h}^2}{\mathrm{me}^2}$$

$$\frac{4 \varepsilon_0^2 n^3 h^3}{m e^4}$$

$$\frac{4 \varepsilon_0^2 \mathrm{n}^2 \mathrm{~h}^3}{m \mathrm{e}^3}$$

MHT CET 2022 11th August Evening Shift

MCQ (Single Correct Answer)

-0

The wave number of the last line of the Balmer series in hydrogen spectrum will be

(Rydberg's constant $$=10^7 \mathrm{~m}^{-1}$$ )

$$250 \mathrm{~m}^{-1}$$

$$2.5 \times 10^6 \mathrm{~m}^{-1}$$

$$0.25 \times 10^9 \mathrm{~m}^{-1}$$

$$2.5 \times 10^5 \mathrm{~m}^{-1}$$

MHT CET 2021 22th September Evening Shift

MCQ (Single Correct Answer)

-0

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)

-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

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

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

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

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

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