91Ó°ÊÓ

The rate of change of the angular velocity with respect to time is known as the angular acceleration.

The expression for energy of electron form Bohr’s theory is given as follows:

$\mathrm{E}=\frac{\left(-13.6\right){\mathrm{Z}}^2\text{eV}}{{\mathrm{N}}^2}$

Here, N represents the orbit number and Z is the atomic number.

For ground state:

Substitute 1 for N and 2 for Z.

$\begin{array}{rcl}{\mathrm{E}}_1& =& \frac{\left(-13.6\right){\left(2\right)}^2\text{eV}}{{\left(1\right)}^2}\\ & =& -54.4\text{eV}\\ & & \end{array}$

Therefore, the ground state energy of the electron in Helium atom is$-54.4\text{eV}$.

For first excited state:

Substitute 2 for N and 2 for Z in the expression

$\begin{array}{rcl}{\mathrm{E}}_1& =& \frac{\left(-13.6\right){\mathrm{Z}}^2\text{eV}}{{\mathrm{N}}^2}\\ {\mathrm{E}}_2& =& \frac{\left(-13.6\right){\left(2\right)}^2\text{eV}}{{\left(2\right)}^2}\\ & =& -13.6\text{eV}\\ & & \end{array}$

Therefore, the energy of the electron in the first excited state is $-13.6\text{eV}$.

When an electron makes transition from the first excited state to ground state, the energy of the emitted photon is given as follows:

$\mathrm{E}={\mathrm{E}}_2-{\mathrm{E}}_1$

Substitute

$\begin{array}{rcl}{\mathrm{E}}_1& =& -54.4\text{eV}\\ {\mathrm{E}}_2& =& -13.6\text{eV}\\ & & \end{array}$

$\begin{array}{rcl}\mathrm{E}& =& \left(-13.6\right)-\left(-54.4\mathrm{ev}\right)\\ & =& 40.8\text{eV}\\ & & \end{array}$

Hence, the energy of emitted photon is$40.8\text{eV}$.