Question

what are the frequency to wavelength of a photon during a transition from n equal to 5 n equal to 2 state in helium+ ion

Answer 1

Hey,

Use Rydberg's formula, Which is,

1/∆ = RH(1/n1^2-n2^2)z^2

∆(lambda) = which is the wavelength of the photon emitted.

Where RH is Rydberg's constant = 1.097 x 10^7 m^-1

n^1 = lower energy state, i.e 2 in this question.

n^2 = higher energy state, i.e 5

Z = atomic number, here Z for hydrogen is 1

On substituting the given values we get,

1/∆ = 1.097 x 10^7(1/2^2–1/5^2)1^2

1/∆ = 1.097x 10^7(1/4–1/25)

1/∆ = 1.097 x 10^7(0.21)

1/∆ = 0.23 x 10^7

On doing reciprocal we get ∆(lambda), i.e the wavelength of the emitted photon.

∆= 4.347 x 10^-7 m

But we require its frequency,

Therefore, frequency = c/∆

(c= speed of light)

3x10^8/4.347x10^-7 = 0.69 x 10^15

Therefore, frequency= 0.69 x 10^15

Answer 2

Hey your answer.

Hope it helps.....

❤#thank you#❤