Answer :
Given:
The energy difference, E=2.4 eV
To find:
The wavelength of the emitted line.
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,
[tex]E=\frac{hc}{\lambda}[/tex]Where h is the Plank's constant and c is the speed of light.
The energy in joules is given by,
[tex]\begin{gathered} E=2.4\times1.6\times10^{-19} \\ =3.84\times10^{-19}\text{ J} \end{gathered}[/tex]On substituting the known values in the equation of the energy,
[tex]\begin{gathered} 3.84\times10^{-19}=\frac{6.626\times10^{-34}\times3\times10^8}{\lambda} \\ \implies\lambda=\frac{6.626\times10^{-34}\times3\times10^8}{3.84\times10^{-19}} \\ =517.7\times10^{-9}\text{ m} \\ =517.7\text{ nm} \end{gathered}[/tex]Final answer:
Thus the wavelength of the line is 517.7 nm