Given:
Wavelength, λ = 616 nm = 616 × 10−9 m
Speed of light, c = 3 × 108 m s−1
Planck’s constant, h = 6.626 × 10−34 J s
(a) Frequency of emission
ν = c / λ
ν = (3 × 108) / (616 × 10−9)
ν = 4.87 × 1014 s−1
(b) Distance traveled by radiation in 30 s
Distance = speed × time
Distance = 3 × 108 × 30
Distance = 9.0 × 109 m
(c) Energy of one quantum
E = hν
E = (6.626 × 10−34) × (4.87 × 1014)
E = 3.23 × 10−19 J
(d) Number of quanta for 2 J energy
Number of quanta, n = Total energy / Energy per quantum
n = 2 / (3.23 × 10−19)
n = 6.19 × 1018
Final Answers:
(a) Frequency = 4.87 × 1014 s−1
(b) Distance traveled = 9.0 × 109 m
(c) Energy of one quantum = 3.23 × 10−19 J
(d) Number of quanta = 6.19 × 1018
Considering Bohr’s atomic model for hydrogen atom :
(A) the energy of H atom in ground state is same as energy of He+ ion in its first excited state.
(B) the energy of H atom in ground state is same as that for Li++ ion in its second excited state.
(C) the energy of H atom in its ground state is same as that of He+ ion for its ground state.
(D) the energy of He+ ion in its first excited state is same as that for Li++ ion in its ground state.