Question

Write the expression for Bohr’s radius in hydrogen atom.

Answer 1

 The expression for Bohr’s radius in hydrogen atom is.

where: = permittivity of free space
 = reduced plank's constant = h/2π 
 = electron rest mass 
 = elementary charge
 =  speed of light in vacuum
 = structural constant.

Answer 2

Answer:

$r=\frac{n^2}{Z} \times 5.292 \times 10^{-11}\; m$

Explanation:

According to Bohr's Theory,  

The force of attraction acting between the electron and the nucleus is equal to the centrifugal force acting on the electron.

Thus,  

$\frac{Ze^2}{4\pi \epsilon_0 r^2}=m_e\frac{v^2}{r}$ ......1

Where, Z is the atomic number or the number of protons

r is the atomic radius

v is the velocity of the electron

$m_e$ is the mass of the electron

Also,  

Accoriding to Bohr, the angular momentum is quantized. He states that the angular momemtum is equal to the integral multiple of $\frac {h}{2\times \pi}$.

$m_e vr=n\times \frac {h}{2\times \pi}$ ....2

solving r from equation 2, we get that:

$v=n\times \frac {h}{2\times \pi\times m_e r}$

Putting in 1 , we get that:

$r=\frac{4\pi \epsilon_0 \hbar^2}{Ze^2 m_e}n^2 \\ &=\dfrac{a_0}{Z}n^2$

Applying values for hydrogen atom,  

Z = 1

Mass of the electron ($m_e$) is 9.1093×10⁻³¹ kg

Charge of electron (e) is 1.60217662 × 10⁻¹⁹ C

$\epsilon_0$ = 8.854×10⁻¹² C² N⁻¹ m⁻²

h is Plank's constant having value = 6.626×10⁻³⁴ m² kg / s

We get that:

$r=\frac{n^2}{Z} \times 5.292 \times 10^{-11}\; m$