Question

What is the wavelength (in nanometers) of a photon emitted during a transition from ni = 7 to nf = 3 state in the H atom?

Answer 1

`\mathbf{\lambda = 1.01 * 10^(3) \ nm}`

Step-by-step explanation:

To determine the wavelength; let's first calculate the energy (E) change of the transition by using the Rydberg constant
`(R_H)`

`\Delta E = R_H ( (1)/(n_i^2)- (1)/(n_f^2))`

`\Delta E = 2. 18 * 10^(-18) \ J ( (1)/(7^2)- (1)/(3^2))`

`\Delta E = 1.977 * 10^(-19 ) \ J`

Now; to calculate the wavelength by using the formula:

`\lambda = (c*h)/(\Delta E)`

`\lambda = ((3.00 * 10^8 \ m/s) *(6.63 * 10^(-34) \ J.s))/(1.977 * 10^(-19 ) \ J)`

`\lambda = 1.01 * 10^(-6) \ m`

To nanometers; we have:

`\lambda = 1.01 * 10^(-6) \ m * (1 \ nm)/( 1 * 10^(-9) \ m)`

`\mathbf{\lambda = 1.01 * 10^(3) \ nm}`