Question

 

An electron in the hydrogen atom falls from the 2p to 1s state and a photon is emitted. What is the wavelength of the emitted photon (in nm)?

Select one:

a. 20

b. 91

c. 122

d. 364

e. 138

Answer 1

If Ei and Ef are the energies of initial and final state of the photon; where initial state corresponds to higher energy and final state corresponds to lower energy so that photon jumps from initial energy state (ni )to final energy state (nf) by emitting the radiation; then the wavelength(k) of the emitted radiation would be given by the formula

1/k = RH((1/nf)2 - (1/ni2)) .............eqn 1

where RH = 1.07678*107 m

As the question says that there is the transistion from 2p to 1s;

we know that the electronic configuration is written in the form of : ns, np etc

Now, if we clearly see the initial state i.e 2p

the value of n = 2 in this case i.e initial energy state is 2

ni = 2

Further, final state is 1s

the value of n = 1 in this case i.e final energy state is 1

nf = 1

Using the value of RH, ni and nf in eqn 1; we get

i.e option c is the correct answer

The wavelength k associated with this transition is 122 nm.

Also this corresponds to Lyman series as the final state, nf = 1 in this case.