Answer:
the wavelength of radiation emitted is [tex]\mathbf{\lambda= 2169.62 \ nm}[/tex]
Explanation:
The energy of the Bohr's hydrogen atom can be expressed with the formula:
[tex]\mathtt{E_n =- \dfrac{13.6\ ev}{n^2}}[/tex]
For n = 7:
[tex]\mathtt{E_7 =- \dfrac{13.6\ ev}{7^2}}[/tex]
[tex]\mathtt{E_7 =-0.27755 \ eV}[/tex]
For n = 4
[tex]\mathtt{E_4=- \dfrac{13.6\ ev}{4^2}}[/tex]
[tex]\mathtt{E_4 =- 0.85\ eV}[/tex]
The electron goes from the n = 7 to the n = 4, then :
[tex]\mathtt{E_7-E_4 = (-0.27755 - (-0.85) ) \ eV}[/tex]
[tex]\mathtt{= 0.57245\ eV}[/tex]
Wavelength of the radiation emitted:
[tex]\mathtt{\lambda= \dfrac{hc}{0.57245 \ eV}}[/tex]
where;
hc = 1242 eV.nm
[tex]\mathtt{\lambda= \dfrac{1242 \ eV.nm }{0.57245 \ eV}}[/tex]
[tex]\mathbf{\lambda= 2169.62 \ nm}[/tex]
