Question

Given the wavelength of the corresponding emission line, calculate the equivalent radiated energy from n = 4 to n = 2 in both joules and electron volts. Also, calculate the frequency of the wave. λ (Á) = 4861, ƒ(Hz), E(J), E(eV)

Answer 1

Explanation:

It is given that,

Initial orbit of electrons, $n_i=4$

Final orbit of electrons, $n_f=2$

We need to find energy, wavelength and frequency of the wave.

When atom make transition from one orbit to another, the energy of wave is given by :

$E=-13.6(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})$

Putting all the values we get :

$E=-13.6(\dfrac{1}{(4)^2}-\dfrac{1}{(2)^2})\\\\E=2.55\ eV$

We know that : $1\ eV=1.6\times 10^{-19}\ J$

So,

$E=2.55\times 1.6\times 10^{-19}\\\\E=4.08\times 10^{-19}\ J$

Energy of wave in terms of frequency is given by :

$E=hf$

$f=\dfrac{E}{h}\\\\f=\dfrac{4.08\times 10^{-19}}{6.63\times 10^{-34}}\\\\f=6.14\times 10^{14}\ Hz$

Also, $c=f\lambda$

$\lambda$ is wavelength

So,

$\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{6.14\times 10^{14}}\\\\\lambda=4.88\times 10^{-7}\ m\\\\\lambda=488\ nm$

Hence, this is the required solution.