Question

What wavelength photon is required to excite a hydrogen from the n=1 state to the n=3 state? What is the lowest frequency photon that will be observed?

Answer 1

1.032 x 10^-7 m, 2.9 x 10^15 Hz

Step-by-step explanation:

n = 1 to n = 3

Rydberg's constant, R = 1.09 × 10^7 per metre

Use the formula for the wavelength

`(1)/(\lambda )=R\left ( (1)/(n_(1)^(2))-(1)/(n_(2)^(2)) \right )`

here, n1 = 1 and n2 = 3

`(1)/(\lambda )=1.09* 10^(7)\left ( (1)/(1^(2))-(1)/(3^(2)) \right )`

`(1)/(\lambda )=1.09* 10^(7)* (8)/(9)`

λ = 1.032 x 10^-7 m

Let the frequency is f.

Use the relation

v = f x λ

`3 * 10^8=f* 1.032 * 10^(-7)`

f = 2.9 x 10^15 Hz

Answer 2

Step-by-step explanation:

It is given that,

Initial state of electron,
`n_i=1`

Final state of electron,
`n_f=3`

The wavelength of the excited electron is given by :

`(1)/(\lambda)=R((1)/(n_f^2)-(1)/(n_i^2))`

Where

R is Rydberg's constant

`(1)/(\lambda)=1.097* 10^(7)\ J* ((1)/(3^2)-(1)/(1^2))`

`\lambda=-1.02* 10^(-7)\ m`

or

`\lambda=102\ nm`

Let f is the frequency of the observed photon. It is given by :

`f=(c)/(\lambda)`

`f=(3* 10^8\ m/s)/(1.02* 10^(-7)\ m)`

`f=2.94* 10^(15)\ Hz`

Hence, this is the required solution.