Question

Alpha particles emitted from radioactive cesium have an energy of 4.8 MeV. Calculate the deBroglie wavelength of one of these particles. Note that the mass of an alpha particle = 6.6 x 10-25 g and h = 6.60 x 10-27 sec.

Answer 1

Ans-

Bohr postulated that an atom will emit radiation only when the electron, initially in one of the 
stable allowed orbits where E = E u , changes to another allowed orbit with a smaller energy given by 
E = E,. The energy of the emitted photon will then be equal to the difference between the electron 
energies in the two allowed orbits. Thus the wavelength of the emitted photon will be found from 

Ey = hv = h^ =E U -E, or I-JL^-zr,) {19J) 

Substituting the values of the orbital energies given by (19.5), we then find 

I _ IvW mZ 2 ( 1 ±) = R z 2 (— ' 



A h\ \nf n 2 J °° \n 2 n 2 

with 

2TT 2 (ke 2 ) 2 (mc 2 ) 
R^= -i = 1.09737 X 10- 3 A-' 



(he) 



CHAP. 19] 



THE BOHR ATOM 



101 



In this analysis it was assumed that the positively charged nucleus is so massive compared to the 
electron that it could be considered infinitely heavy. If the finite mass of the nucleus is taken into 
consideration, the motion of the combined electron (m) and nucleus (M) system, separated by distance 
r, about its center of mass (mp =MP, r =p +P) is equivalent to a particle of reduced mass 

m M 



1 + -2- 
M 



1 + M 
m 



orbiting the center of mass at a radius r. For hydrogen m/M = 1/1836, and using this to modify the 
Rydberg constant we obtain 

R «> 1.09737 X 10~ 3 A" 



R " 1 + (m/M) 



1 + (1/1836) 
in agreement with the experimental value R = 1.0967758 X 10" 3 A 



= 1.0968 X 10- 3 A"