Given:
Wavelength of incident radiation, λ = 6800 Å = 6.8 × 10−7 m
Speed of light, c = 3.0 × 108 m s−1
Planck’s constant, h = 6.626 × 10−34 J s
Concept Used:
Since electrons are emitted with zero velocity, the given wavelength corresponds to the threshold wavelength.
Step 1: Calculate threshold frequency (ν0)
ν0 = c / λ
ν0 = (3.0 × 108) / (6.8 × 10−7)
ν0 = 4.41 × 1014 s−1
Step 2: Calculate work function (W0)
W0 = hν0
W0 = (6.626 × 10−34) × (4.41 × 1014)
W0 = 2.92 × 10−19 J
Step 3: Convert work function into electron volts
1 eV = 1.602 × 10−19 J
W0 = (2.92 × 10−19) / (1.602 × 10−19)
W0 = 1.82 eV
Final Answer:
Threshold frequency, ν0 = 4.41 × 1014 s−1
Work function of the metal, W0 = 2.92 × 10−19 J (≈ 1.82 eV)
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.