Question

Find wavelength of photon emitted during its transition from 4E/3 level to E level, if $\lambda$ is the wavelength emitted during transition from 2E level to E level in the following diagram :- (Here E represents the energy of shell)

• A. $\lambda/3$
• B. $3\lambda/4$
• C. $4\lambda/3$
• D. $3\lambda$

Answer 1

Answer: (D)

3λ

Answer 2

To find the wavelength of the emitted photon, we use the relationship between energy difference and wavelength:

$E = \frac{hc}{\lambda}$

where $E$ is the energy of the photon, $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength.

Given:

1. A photon of wavelength $\lambda$ is emitted during a transition from the 2E level to the E level. The energy difference for this transition ($\Delta E_1$) is: $\Delta E_1 = 2E - E = E$ So, we can write: $E = \frac{hc}{\lambda}$ (Equation 1)

2. We need to find the wavelength (let's call it $\lambda'$) of the photon emitted during the transition from the 4E/3 level to the E level. The energy difference for this transition ($\Delta E_2$) is: $\Delta E_2 = \frac{4E}{3} - E = \frac{4E - 3E}{3} = \frac{E}{3}$ So, we can write: $\frac{E}{3} = \frac{hc}{\lambda'}$ (Equation 2)

Now, we need to express $\lambda'$ in terms of $\lambda$. From Equation 1, we can express $hc$: $hc = E\lambda$

Substitute this expression for $hc$ into Equation 2: $\frac{E}{3} = \frac{E\lambda}{\lambda'}$

Since E is a non-zero energy value, we can cancel E from both sides of the equation: $\frac{1}{3} = \frac{\lambda}{\lambda'}$

Now, solve for $\lambda'$: $\lambda' = 3\lambda$

Thus, the wavelength of the photon emitted during the transition from the 4E/3 level to the E level is $3\lambda$.