1.

Read the pasage given below as well as theadjoining energylevel diagram and then answer the questions given afterthe passage . As per Bohr'stheory oghydrogen atom the energy of an atom in a statecorresponding to principal quantum numbern is given as : E_(n) = - (13.6)/(n^(2)) eV Theenergyof an atom is the least when itselectron is revolvingin an orbit closets to thenucleus i.e.,theone forwhich n =1. This stateis called the groundstate. When a hydrogen atom receives energyby processes suchas electron collision, theatom may acquire sufficientenergyto raise theelectron to higherenergystates and the atom is then saidto bein an excited state . From theseexcited states the electron can then fall back to a state of lower energy , emitting a photon in the process. The energy level diagram for thestationary states of a hydrogen atom , compouted from Bohr'srelation for energy, is givenin Fig.12.02. Theprincipalquantum number n labels the stationary states in the ascendingorderof energy . Obviously , thehighestenergycorrespondsto n = ooandhasan energy of 0 eV. Thisis the energy of the atom when the electron is completely removed (r= oo) from the nucleus and is at rest. Now answer the follwoingquestions: Calculate the wavelenghtof firstspectral lines of series mentioned in part (g) of question.

Answer»

SOLUTION :Firstline of Balmer seriescorresponds totransition from `n_(1) = 3` to `n_(G) = 2 ` . Hence, we have
`(h_(c))/(lambda_(1)) =E_(3) - E_(2) = (-1.51) - (-3.40) = 1.89 eV`
`rArr""lambda_(1) = (h_(c))/(1.89eV) = ((6.63 XX 10^(-34)) xx (3XX10^(8)))/(1.89 xx(1.6 xx 10^(-19))) = 6.577 xx 10^(-7) m = 657.7nm`


Discussion

No Comment Found

Related InterviewSolutions