Question

# The ground state energy of hydrogen atom is 13.6 eV. When its electron is in the first excited state, its excitation energy is:

A
10.2 eV
B
zero
C
3.4 eV
D
6.8 eV

Solution

## The correct option is B 10.2 eVThe energy of an electron in Bohr’s orbit of hydrogen atom is given by the expression:$$E_n=-13.6 \dfrac{Z^2}{n^2}$$; [where n = energy state of electron]For hydrogen Z = 1$$E_1= -13.6 eV$$ (given)For the first excited state, n = 2,Therefore, $$E_2=\dfrac{-13.6}{2^2} eV$$ $$E_2= -3.4\ eV$$Now, $$\Delta{E}=E_2-E_1$$ $$= -3.4 - (-13.6) = 10.2 eV$$Chemistry

Suggest Corrections

0

Similar questions
View More

People also searched for
View More