Figure below shows variation of stopping potential (V0) with frequency

1.	Figure below shows variation of stopping potential (V0) with frequency(?) of incident radiations for two different metals A and B.1. Write down the values of work function A and B.2. What is the significance of slope of the above graph?3. The value of stopping potential for A and B for a frequency γ01 (which is greater than γ02) of incident radiations are V1 and V0 respectively. Show that the slopes of the lines is equal to \(\frac{v_1-v_2}{γ_{01}-γ_{02}}\).
Answer» 1. Work function of A, Φ₀₁ – hν₀₁. Work function of B, Φ₀₁ – hν₀₂. 2. The slope of the graph gives value of h/e. 3. For the metal A, hν₁ = hν₀₁ + eV₁ ……(1) For the metal B, hν₁ = hν₀₂ + eV₂ ……..(2) From equation (1) and (2) hν₀₁ + eV₁ = hν₀₂ + eV₂ e(V₁ – V₂) = h(ν₀₂ – ν₀₁) \(\frac{h}{e} = \frac{V_1-V_2}{v_{02}-v_{01}}\)

Figure below shows variation of stopping potential (V0) with frequency(?) of incident radiations for two different metals A and B.1. Write down the values of work function A and B.2. What is the significance of slope of the above graph?3. The value of stopping potential for A and B for a frequency γ01 (which is greater than γ02) of incident radiations are V1 and V0 respectively. Show that the slopes of the lines is equal to \(\frac{v_1-v_2}{γ_{01}-γ_{02}}\).

Answer»

1. Work function of A, Φ₀₁ – hν₀₁.

Work function of B, Φ₀₁ – hν₀₂.

2. The slope of the graph gives value of h/e.

3. For the metal A, hν₁ = hν₀₁ + eV₁ ……(1)

For the metal B, hν₁ = hν₀₂ + eV₂ ……..(2)

From equation (1) and (2)

hν₀₁ + eV₁ = hν₀₂ + eV₂

e(V₁ – V₂) = h(ν₀₂ – ν₀₁)

\(\frac{h}{e} = \frac{V_1-V_2}{v_{02}-v_{01}}\)