In experimenting with rubidium photocell, the following lines from a mercury source were used λ1=3650A0, λ2=4047A0, λ3=4358A0, λ4=5461A0, λ5=6907A0 and stopping potentials respectively are V01 = 1.28V, V02 = 0.95V, V03 = 0.74V, V04 = 0.16V, V05 = 0V. Threshold frequency and work function of metal are

E=W0+KmaxE=W0+eV124003650=W0+1.28W0=3.39−1.28=2.11eVν0=W0h=5×1014 Hz

In experimenting with rubidium photocell, the following lines from a mercury source were used λ1=3650A0, λ2=4047A0, λ3=4358A0, λ4=5461A0, λ5=6907A0 and stopping potentials respectively are V01 = 1.28V, V02 = 0.95V, V03 = 0.74V, V04 = 0.16V, V05 = 0V. Threshold frequency and work function of metal are

E=W0+KmaxE=W0+eV124003650=W0+1.28W0=3.39−1.28=2.11eVν0=W0h=5×1014 Hz

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In experimenting with rubidium photocell, the following lines from a mercury source were used $λ_{1} = 3650 \overset{0}{A}, λ_{2} = 4047 \overset{0}{A}, λ_{3} = 4358 \overset{0}{A}, λ_{4} = 5461 \overset{0}{A}, λ_{5} = 6907 \overset{0}{A}$ and stopping potentials respectively are V₀₁ = 1.28V, V₀₂ = 0.95V, V₀₃ = 0.74V, V₀₄ = 0.16V, V₀₅ = 0V. Threshold frequency and work function of metal are

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1.5 × 10¹⁴Hz; 5eV

4 × 10¹⁴Hz; 1.5eV

4 × 10¹⁴Hz; 3eV

5 × 10¹⁴Hz; 2.1eV

answer is B.

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Detailed Solution

$E = W_{0} + K_{max}$

$E = W_{0} + eV$

$\frac{12400}{3650} = W_{0} + 1.28$

$W_{0} = 3.39 - 1.28 = 2.11 eV$

$ν_{0} = \frac{W_{0}}{h} = 5 \times 10^{14} Hz$

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