Given:
Power of bulb, P = 25 W = 25 J s−1
Wavelength of light, λ = 0.57 µm = 0.57 × 10−6 m
Speed of light, c = 3.0 × 108 m s−1
Planck’s constant, h = 6.626 × 10−34 J s
Step 1: Calculate energy of one quantum (photon)
E = hc / λ
E = (6.626 × 10−34 × 3.0 × 108) / (0.57 × 10−6)
E = 3.49 × 10−19 J
Step 2: Calculate rate of emission of quanta
Rate of emission = Power / Energy per quantum
Number of quanta per second = 25 / (3.49 × 10−19)
= 7.17 × 1019 quanta s−1
Final Answer:
The rate of emission of quanta is:
7.17 × 1019 photons per second
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.