Question

Which of the following electron transitions in a hydrogen atom results in the emission of photons with the shortest wavelength? A) n = 3 to n = 4 B) n = 1 to n = 3 C) n = 6 to n = 4 D) n = 7 to n = 5 E) n = 2 to n = 5

Answer 1

if energy is more the corresponding wavelength will be shortest.

so that

answer:

E = hv   = hcV

V =wavenumber

A) = (6.625*10^(-34)*3*10^8)*(109677*(1/n1^2 - 1/n2^2)    (n2>n1)

    = (6.625*10^(-34)*3*10^8)*(109677*((1/3^2) - (1/4^2))   = 1.05*10^-21 j

B) E = (6.625*10^(-34)*3*10^8)*(109677*(1/n1^2 - 1/n2^2)    (n2>n1)

    = (6.625*10^(-34)*3*10^8)*(109677*((1/1^2) - (1/3^2))   = 1.937627*10^-20 j

c)    = (6.625*10^(-34)*3*10^8)*(109677*((1/4^2) - (1/6^2))   = 7.57*10^-22 j

d)   = (6.625*10^(-34)*3*10^8)*(109677*((1/5^2) - (1/7^2))   = 4.27*10^-22 j

e)   = (6.625*10^(-34)*3*10^8)*(109677*((1/2^2) - (1/5^2))   = 4.58*10^-21 j

highest energy   =   B