Question 478: Atoms & Nuclei

Question

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)

Accepted Answer

The velocity of an electron in the first orbit of a hydrogen atom according to Bohr's model can be derived from the quantization of angular momentum and the electrostatic force balance with centripetal force. The velocity $v_n$ of an electron in the nth orbit is given by:
$v_n = \frac{e^2}{2 \varepsilon_0 h} \cdot \frac{1}{n}$
For the ground state (first orbit, $n=1$), the velocity $v_1$ is:
$v_1 = \frac{e^2}{2 \varepsilon_0 h}$
To find the ratio of this velocity to the speed of light $c$, we divide the expression for $v_1$ by $c$:
$\frac{v_1}{c} = \frac{\frac{e^2}{2 \varepsilon_0 h}}{c}$
$= \frac{e^2}{2 \varepsilon_0 h c}$
Therefore, the correct option is:
Option C: $\frac{e^2}{2 \varepsilon_0 h c}$

Answer

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

Answer

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

Answer

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

Answer

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

Atoms & Nuclei

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)

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