DUAL NATURE OF MATTER AND RADIATION

Answer :

given: ɸ = 4 eV = 4 × 1.6 × 10^–19 J

                             ɸ = (hc / λ)

                   hence λ = (hc / ɸ)

                             λ = [{6.62 × 10^–34 × 3 × 10⁸} / {4 × 1.6 × 10^–19}]

                                = 3.103 × 10^–7 m

                                = 3.103 × 10^–9 × 10² m

                                = 310.3 nm   

Answer :

m = 200 gm = 0.2 kg

                             V = 5 (m / br) = {5 / (3600)} m/s

                             P = mV = (h/λ)

                             ∴ λ = (h/p) = (h / mV)

                             ∴ λ = [{6.62 × 10^–34} / {0.2 × {5 / (3600)}}]

                                   = 23.83 × 10³ × 10^–34

                             ∴ λ = 2.38 × 10^–30 m i.e. order is 10^–30 m   

Answer :

Kinetic energy = E = (1/2) mV² = hf – hf_o

                             ∴ E = h(f – f_o)

                   ∴ (E₁ / E₂) = [{h(f₁ – f_o)} / {h(f₂ – f_o)}] = {(f₁ – f_o) / (f₂ – f_o)}             (1)

                   given: f₁ = 5.5 × 10⁸ MHz = 5.5 × 10¹⁴ Hz

                             f₂ = 4.5 × 10⁸ MHz = 4.5 × 10¹⁴ Hz

                   also (E₁ / E₂) = (1/5)

                                       ∴ (1/5)        = {(f₁ – f_o) / (f₂ – f_o)} ----- from (1)

                                      ∴ f₂ – f_o        = 5f₁ – 5f_o

                                    ∴ f₂ – 5f₁        = – 4 f_o

                   ∴ f_o = Threshold frequexy = {(5f₁ – f₂) / 4} = [{5(5.5 × 10¹⁴)}  / 4]  – 4.5 × 10¹⁴

                             ∴ f_o = [{23 × 10¹⁴} / 4] = 5.75 × 10¹⁴ Hz

                                f_o = 5.75 × 10⁸ MHz   

Answer :

λ = (h/P)                                                               (1)

                   λ change by 0.25% then momentum changes by P_O.

                   hence

                             λ [1 + {(0.25) / (100)}] = {h / (P – P_O)}                                   (2)

                   ∴ ratio of (1) of (1) and (2) gives

                             {(100) / (100.25)} = {(P – P_O) / P}

                                           ∴ 100P = 100.25P – 100.25 P_O

                                     ∴ 100.25 P_O = 0.25P

                                                ∴ P = 401 P_O

                   i.e. initial momentum is 401P_O 

Answer :

graph: Kinetic energy versus frequency of radiation. graph is given to be lines. Meaning that its slope is same for all metals and free from intensity of radiation.