1. With increases in principal quantum number n the energy difference between adjacent energy
levels in hydrogen atom
1) increases 2) decreases 3) remain constant
4) decreases for lower values of n and increases for higher values of n
Ans: 2
4) decreases for lower values of n and increases for higher values of n
Ans: 2
Sol: N
EN = --13.6/n2 ev
As value of n increases the energy difference between adjacent levels decreases.
will be
1) 25 : 9 2) 5 : 3 3) 9 : 25 4) 3 : 5
Ans: 1
Sol: According to de-Broglie equation
ƛ=h/nv but ƛ1 : ƛ2 =3:5
.'. v1:v2=5:3
K.E = ½ mv2
K.E1=K.E2
= 52 : 32= 25: 9
3. The velocities of two particles A and B are 0.05 and 0.02 ms–1 respectively. The mass of B is five
times the mass of A. The ratio of their de-Broglie’s wavelength is (E-2008)
1) 2 : 1 2) 1 : 4 3) 1 : 1 4) 4 : 1
Ans: 1
Sol: ƛ=h/mv
ƛA=h/m*0.05 ; ƛB = ƛ/5m*0.02
ƛA/ ƛB = 5m*0.02/m*0.05 = 2 : 1
4.The wavelength (in A°) of an emission line obtained for Li2+ during electronic transition from
n2 = 2 to n1= 1 is (R = Rydberg constant)
1)3R/4 2)27R/4 3) 4/3R 4) 4/27R
ANS: 4
sol: for Li2+ = 32R[1/12 –1/22] = 27R/4
ƛ = 4/27R
5.An electronic transition in hydrogen atom results in the formation of Hα line of hydrogen in
Lyman series, the energies associated with the electron in each of the orbits involved in the
transition (in kcal mol–1) are
1) −313.6,−34.84 2) −313.6 ,− 78.4 3) −78.4,−34.83 4) −78.4,−19.6
Ans: 2
Sol: Hα line in Lyman series mean electron transition is from n = 2 to n = 1 orbit.
E1 = --313.6/12= 313.6 313.6 k.calmole-1
E2 = --313.6/22= --78.4 k.calmole-1
