Answer:[tex]1.08\times 10^{-8}m[/tex]
Explanation:
Let [tex]E_{n}[/tex] be the energy of the electron in [tex]n[/tex]th orbit.
According to Bohr's model,
[tex]E_{n}=\frac{-kz^{2}}{n^{2}}[/tex]
where [tex]k=2.179\times 10^{-18}J[/tex]
[tex]Z[/tex] is the atomic number
[tex]n[/tex] is the orbit number.
Given,[tex]z=3[/tex]
Energy required for transition from [tex]n=1[/tex] to [tex]n=4[/tex] is [tex]\frac{k(3)^{2}}{1^{2}}-\frac{k(3)^{2}}{4^{2}}=\frac{15k\times 9}{16} =18.38\times 10^{-18}J[/tex]
Since,wave length is [tex]\frac{hc}{E}[/tex]
where [tex]h[/tex] is the plancks constant.
[tex]c[/tex] is the speed of light.
[tex]c=3\times 10^{8}\\h=6.63\times 10^{-34}m^{2}KgS^{-1}[/tex]
So,wave length is [tex]\frac{6.63\times 10^{-34} \times 3\times 10^{8}}{18.38\times 10^{-18}} =1.08\times 10^{-8}m[/tex]
