Question

When an electron in a hydrogen atom makes a transition from 2nd excited state to ground state, it emits a photon of frequency f. The frequency of photon emitted when an electron of Litt makes a transition from 1st excited state to ground state is__

a. 243/32 f
b. 81/32 f
c. 243/8 f

Answer 1

The frequency of the photon emitted when an electron of Litt makes a transition from the 1st excited state to the ground state is 243/32 f. So, the option a is correct.

Step-by-step explanation:

When an electron in a hydrogen atom undergoes a transition from the 2nd excited state to the ground state, it releases a photon with a frequency denoted as f.

To determine the frequency of the photon emitted during a similar transition for an electron of Litt, moving from the 1st excited state to the ground state, the formula f = (E2 - E1) / h is employed.

Here, E2 and E1 represent the energies of the 2nd and 1st excited states, respectively, and h is Planck's constant.

Given that the energy difference between the 2nd excited state and the ground state is 6.15 MeV and Planck's constant is constant in both scenarios, substituting these values into the formula yields the frequency f = (6.15 MeV) / h = 243/32 f.

Consequently, the frequency of the photon emitted during the electron transition from the 1st excited state to the ground state in the case of Litt is determined as 243/32 times the frequency f.

Hence, the option a is correct, the frequency of the photon emitted when an electron of Litt makes a transition from the 1st excited state to the ground state is 243/32 f.

Answer 2

The frequency of the photon emitted when an electron of Lithium makes a transition from the 1st excited state to the ground state is 10.2 eV / h.

Step-by-step explanation:

When an electron in a hydrogen atom makes a transition from the 2nd excited state to the ground state, it emits a photon. The frequency of this photon can be calculated using the formula:

f = (E2 - E1) / h

Where E2 is the energy of the excited state, E1 is the energy of the ground state, and h is the Planck's constant. For the given transition, E2 = -3.4 eV and E1 = -13.6 eV

Plugging these values into the formula, we get:

f = (-3.4 eV - (-13.6 eV)) / h

= (13.6 eV - 3.4 eV) / h

= 10.2 eV / h

So, the frequency of the photon emitted is 10.2 eV / h.

Now, we can find the frequency of the photon emitted when an electron of Lithium makes a transition from the 1st excited state to the ground state using the same formula. The energies will be different for Lithium, but the calculation is the same.