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Masteriza [31]
3 years ago
14

Given the wavelength of the corresponding emission line, calculate the equivalent radiated energy from n = 4 to n = 2 in both jo

ules and electron volts. Also, calculate the frequency of the wave. λ (Á) = 4861, ƒ(Hz), E(J), E(eV)
Chemistry
1 answer:
lions [1.4K]3 years ago
4 0

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.

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