Question

Find the temperature at which the average thermal kinetic energy is equal to the energy needed to take a hydrogen atom from its ground state to n = 3 state. Hydrogen can now emit red light of wavelength 653.1 nm. Because of Maxwellian distribution of speeds, a hydrogen sample emits red light at temperatures much lower than that obtained from this problem. Assume that hydrogen molecules dissociate into atoms.

Answer 1

The temperature at which the thermal kinetic Energy of an hydrogen atom is sufficient to excite an electron in a ground state to an excited state so that it emits a red photon of wavelength 653.1nm can be calculated using the black body equation which relates energy with temperature.

1)According to the quantum equation hv=5KT where h is equal to planks constant,v is photon frequency ,K is a constant and t is temperature in Kelvin .

2)First we will calculate photon frequency from the wavelength using formula c=f lambda where c is speed of light and lambda is the wavelength which is 653.1nm.

3)We get a frequency of 0.472×10^15 Hertz.

4)It is important to note down that the wavelength given is in nm so we have to convert it to metred before measuring frequency .

5)once we get the frequency we just have to use it in the above equation.

6)The value for k is 1.38×10^-23 joules.

7)After applying the equation hv=5KT we get a temperature of 4530 Kelvin which corresponds to a temperature of 4256.85 Celsius.

8)At this temperature the kinetic exceeds to excite an electron from the griubd state to the excited state so that a red photon is released .

Answer 2

The temperature at which the average thermal kinetic energy is equal to the energy needed to take a hydrogen atom from its ground state to n = 3 state is $9.4 \times 10^{4} K$

Explanation:

In the question, it is given:

Red light has the wavelength, λ = 653.1 nm

$H_2$ molecules has the kinetic energy, K that is shown as K = 32 kT   …(1)

where, T = $H_2$ molecule's temperature,$k = 8.62 \times10^{-5} eV/K$

When the atom goes to n = 3 from ground state, the released energy is shown as

$E = 13.6(\frac{1}{n_12} - \frac{1}{n_22})$

For ground state, $n_2$= 3 and $n_1$= 1

$E = 13.6 (\frac{1}{1^2} - \frac{1}{3^2} )$

$E = 13.6 (\frac{1}{1} - \frac{1}{9} )$

Therefore,$E = 13.6 (\frac{8}{9} )$…(2)

$H_2$ molecules has the kinetic energy = When hydrogen atom goes to n = 3 state from ground state, the released energy,

Therefore, $\frac{3}{2} \times 8.62\times T = 13.6\times \frac{8}{9}$

⇒ $T = \frac{8 \times 13.6 \times 2}{3 \times 8.62 \times 9}$

$T = 9.4 \times 10^4K$