The surface of some material is illuminated by waves of λ1=3.5×10−7m and λ2=5.4×10−7m respectively. The ratio of the stopping potential in two cases is 2:1. The work function of the material is

E=ω0+KE1=123753500=3.54E2=123755400=2.2921=3.54−w02.29−w0w0=1.04 ev

The surface of some material is illuminated by waves of λ1=3.5×10−7m and λ2=5.4×10−7m respectively. The ratio of the stopping potential in two cases is 2:1. The work function of the material is

E=ω0+KE1=123753500=3.54E2=123755400=2.2921=3.54−w02.29−w0w0=1.04 ev

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The surface of some material is illuminated by waves of $λ_{1} = 3.5 \times 10^{- 7} m$ and $λ_{2} = 5.4 \times 10^{- 7} m$ respectively. The ratio of the stopping potential in two cases is 2:1. The work function of the material is

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1.04 eV

1.34 eV

2.20 eV

0.86 eV

answer is C.

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Detailed Solution

$E = ω_{0} + K$

$E_{1} = \frac{12375}{3500} = 3.54$

$E_{2} = \frac{12375}{5400} = 2.29$

$\frac{2}{1} = \frac{3.54 - w_{0}}{2.29 - w_{0}}$

$w_{0} = 1.04 e v$

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The surface of some material is illuminated by waves of λ1=3.5×10−7m and λ2=5.4×10−7m respectively. The ratio of the stopping potential in two cases is 2:1. The work function of the material is

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The surface of some material is illuminated by waves of $λ_{1} = 3.5 \times 10^{- 7} m$ and $λ_{2} = 5.4 \times 10^{- 7} m$ respectively. The ratio of the stopping potential in two cases is 2:1. The work function of the material is