MHT CET 2020 16th October Morning Shift | Atoms and Nuclei Question 86 | Physics

The ratio of speed of an electron in the ground state in the Bohr's first orbit of hydrogen atom to velocity of light $$(c)$$ is ( $$h=$$ Planck's constant, $$\varepsilon_0=$$ permittivity of free space, $$e=$$ charge on electron)

$$\frac{2 \theta^2 \varepsilon_0}{h c}$$

$$\frac{e^3}{2 \varepsilon_0 h c}$$

$$\frac{e^2}{2 \varepsilon_0 h c}$$

$$\frac{2 \varepsilon_0 h c}{e^2}$$

MHT CET 2020 16th October Morning Shift

MCQ (Single Correct Answer)

-0

The force acting on the electrons in hydrogen atom (Bohr's theory) is related to the principle quantum number $$n$$ as

$$n^{-4}$$

$$n^4$$

$$n^{-2}$$

$$n^2$$

MHT CET 2019 3rd May Morning Shift

MCQ (Single Correct Answer)

-0

If the speed of an electron of hydrogen atom in the ground state is $2.2 \times 10^6 \mathrm{~m} / \mathrm{s}$, then its speed in the third excited state will be

$5.5 \times 10^6 \mathrm{~m} / \mathrm{s}$

$5.5 \times 10^5 \mathrm{~m} / \mathrm{s}$

$8.8 \times 10^5 \mathrm{~m} / \mathrm{s}$

$6.8 \times 10^6 \mathrm{~m} / \mathrm{s}$

MHT CET 2019 3rd May Morning Shift

MCQ (Single Correct Answer)

-0

In hydrogen emission spectrum, for any series, the principal quantum number is $n$. Corresponding maximum wavelength $\lambda$ is ( $R=$ Rydberg's constant)

$\frac{R(2 n+1)}{n^2(n+1)}$

$\frac{n^2(n+1)^2}{R(2 n+1)}$

$\frac{n^2(n+1)}{R(2 n+1)}$

$\frac{R(2 n+1)}{n^2(n+1)^2}$

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