Question

What is the wavelength of radiation emitted when an electron goes from the n = 7 to the n = 4 level of the Bohr hydrogen atom? Give your answer in nm.

Answer 1

Answer:

the wavelength of radiation emitted  is $\mathbf{\lambda= 2169.62 \ nm}$

Explanation:

The energy of the Bohr's hydrogen atom can be expressed with the formula:

$\mathtt{E_n =- \dfrac{13.6\ ev}{n^2}}$

For n = 7:

$\mathtt{E_7 =- \dfrac{13.6\ ev}{7^2}}$

$\mathtt{E_7 =-0.27755 \ eV}$

For n = 4

$\mathtt{E_4=- \dfrac{13.6\ ev}{4^2}}$

$\mathtt{E_4 =- 0.85\ eV}$

The  electron goes from the n = 7 to the n = 4, then :

$\mathtt{E_7-E_4 = (-0.27755 - (-0.85) ) \ eV}$

$\mathtt{= 0.57245\ eV}$

Wavelength of the radiation emitted:

$\mathtt{\lambda= \dfrac{hc}{0.57245 \ eV}}$

where;

hc  = 1242 eV.nm

$\mathtt{\lambda= \dfrac{1242 \ eV.nm }{0.57245 \ eV}}$

$\mathbf{\lambda= 2169.62 \ nm}$