Question

The energy of electron in first bohr orbit of the hydrogen atom is -13.6ev ,what is energy of electron in its 2nd bohr orbit-

Answer 1

Answer:

• Energy (E₂) of Electron is - 3.4 eV.

Given:

1. Energy of Electron in the First Orbit = - 13.6 eV.

Explanation:

$\rule{300}{1.5}$

We Know, From Energy of Electron Expression.

$\large\star \: {\boxed{\bold{E_n = - 13.6 \Bigg[\dfrac{Z^2}{n^2}\Bigg] \: eV}}}$

$\bold{Here}\begin{cases}\text{Z Denotes Atomic number} \\ \text{n Denotes Bohr Orbit} \\ \sf{E_n} \: \text{Denotes Energy of nth Shell}\end{cases}$

$\large{\boxed{\tt E_n = - 13.6 \Bigg(\dfrac{Z^2}{n^2}\Bigg)}}$

• For Hydrogen Atomic Number (Z) = 1

Substituting the values,

$\large{\tt \hookrightarrow E_2 = - 13.6 \times \Bigg[\dfrac{(1)^2}{(2)^2}\Bigg]}$

$\large{\tt \hookrightarrow E_2 = - 13.6 \times \Bigg[\dfrac{1}{2}\Bigg]^2}$

$\large{\tt \hookrightarrow E_2 = - 13.6 \times \Bigg[\dfrac{1}{4}\Bigg]}$

$\large{\tt \hookrightarrow E_2 = - 13.6 \times \dfrac{1}{4}}$

$\large{\tt \hookrightarrow E_2 = \cancel{- 13.6} \times \dfrac{1}{\cancel{4}}}$

$\huge{\boxed{\boxed{\tt E_2 = - 3.4 \: eV }}}$

∴ Energy (E₂) of the Second Bohr orbit is - 3.4 eV.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Additional information:-

The Energy of Stationary state of Hydrogen atom is given By:-

$\large\star \: {\boxed{\tt E_n = - R_H \Bigg[\dfrac{1}{n^2}\Bigg]}}$

Here,

• $\large{\sf R_H}$ Denotes Rydbergs Constant.
• n has a variable value I.e. n = 1 , 2 , 3 .....
• The units will be in Joules.

For Stationary state, n = 1

$\large{\tt E_1 = - R_H \Bigg[\dfrac{1}{(1)^2}\Bigg]}$

$\large{\underline{\boxed{\tt E_1 = - R_H \: J}}}$

For Stationary state, n = 2

$\large{\tt E_2 = - R_H \Bigg[\dfrac{1}{(2)^2}\Bigg]}$

$\large{\underline{\boxed{\tt E_2 = \dfrac{- R_H }{4}\: J}}}$

$\rule{300}{1.5}$