Question

A spectral line is emitted when an atom undergoes transition between two levels with a difference in energy of 2.4eV. What is the wavelength of the line?

Answer 1

Given:

The energy difference, E=2.4 eV

To find:

The wavelength of the emitted line.

Step-by-step explanation:

The energy of the wave emitted will be equal to the energy difference between the two energy levels.

The energy of the wave is given by the equation,

`E=(hc)/(\lambda)`

Where h is the Plank's constant and c is the speed of light.

The energy in joules is given by,

`\begin{gathered} E=2.4*1.6*10^(-19) \\ =3.84*10^(-19)\text{ J} \end{gathered}`

On substituting the known values in the equation of the energy,

`\begin{gathered} 3.84*10^(-19)=(6.626*10^(-34)*3*10^8)/(\lambda) \\ \implies\lambda=(6.626*10^(-34)*3*10^8)/(3.84*10^(-19)) \\ =517.7*10^(-9)\text{ m} \\ =517.7\text{ nm} \end{gathered}`

Final answer:

Thus the wavelength of the line is 517.7 nm