Question

A parallel beam of light of wavelength 100 nm passes through a sample of atomic hydrogen gas in ground state. (a) Assume that when a photon supplies some of its energy to a hydrogen atom, the rest of the energy appears as another photon. Neglecting the light emitted by the excited hydrogen atoms in the direction of the incident beam, what wavelengths may be observed in the transmitted beam? (b) A radiation detector is placed near the gas to detect radiation coming perpendicular to the incident beam. Find the wavelengths of radiation that may be detected by the detector.

Answer 1

Answer:

1+1+1+1+1+1+11+1+1+1+1+1+1+1+1+1+11+1+1+131+1+/1+1+1+1+1+131+1+131+1+131+

Answer 2

(a) In the transmitted beam, the observed wavelengths are 560 n-m, 105 n-m and 3881 n-m

(b)  The radiation wavelengths noticed are 121 n-m, 103 n-m and 1911 n-m.

Explanation:

It given that the light’s wavelength, λ = 100 nm  

Incident light’s energy (E) is given by E=hcλ

where,

h = Constant of Planck

λ = Light’s wavelength

Therefore, E = 12.42 eV(a)

Let E2 and E1 be the energies of the 2nd and the 1st state, respectively.

Let the changeover happen to E2 from E1.

Energy captivated (E‘) is shown as

$E' = 13.6( \frac{1}{n_1^2} - \frac{1}{n_2^2})$

E' = 10.2 eV  

Energy left = − 10.2 eV + 12.42 eV = 2.22 eV

Photon’s energy = hcλ

(a) Wavelengths observed in the transmitted beam

We obtain the following by associating the left energy with that of the photon, 2.22 eV = 1242λ or λ = 559.45 = 560 nm

Let the third state’s energy be $E_3$.

Energy engrossed for the changeover to $E_3$ from $E_1$ is shown as

$E' = 13.6( \frac{1}{n_1^2} - \frac{1}{n_2^2})$

E' =89 x 13.6

E' = 12.1 eV

In the transition from $E_1$ to $E_3$, the absorbed energy = 12.1 eV

Energy left = − 12.1 + 12.42 = 0.32 eV

0.32 = hcλ = 12420.32 = 3881 nm

The fourth state’s energy be $E_4$.

In the transition from $E_3$ to$E_4$  the absorbed energy is given by

$E' = 13.6( \frac{1}{n_1^2} - \frac{1}{n_2^2})$

E' = 7144 x 13.6

E' = 0.65 eV

For the transition from n = 3 to n = 4, the energy absorbed is 0.65 eV.

Energy left = 11.77 eV

When this energy is equated with photon’s energy, we obtain                   11.77 = hcλ  

(or) λ = 105.52

In the transmitted beam, the observed wavelengths are 560 n-m, 105 n-m and 3881 n-m.

(b)  Wavelengths of the radiation that may be detected by the detector :

When the absorbed energy by the ‘H’ atom is perpendicularly radiated, then the radiations’ wavelengths noticed are measured in the below method:

E = 10.2 e V = 1910.76 n-m

So, the radiation wavelengths noticed are 121 n-m, 103 n-m and 1911 n-m.