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**

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