What is the wavelength (in nanometers) of a photon emitted during a transition from ni = 7 to nf = 2 state in the H atom?

Respuesta :

Answer:

3.96193 × 10^-7m

Explanation:

I have added the formula used as an attachment.

Now, lambda is the wavelength of the emitted photon

R is the Rydberg's constant

n(final) is the final energy level which in your case is 2

n(initial) is the initial energy level which in your case is 7.

Thus, we input these values into the equation.

1/lambda = 1.0974 × 10^7 ( 1/2^2 - 1/7^2)

1/lambda = 1.0974 × 10^7 × 0.23

= 2,524,020m^-1

Hence lamba = 1/2524020 =

3.96193 × 10^-7m

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