For intensity I of a light of wavelength 5000 A the photoelectron saturation current is 0-40 - A and stopping potential is 1-36 V- the work function of metal isA- 2-47 eVB- 1-36 eVC- 1-10 eVD- 0-43 eV

The correct option is C 1.10 eV By using E=W_0+K_max E =12375/5000= 2.475 eV and K_max= eV_0=1.36eV So 2.475=W_0 + 1.36 ⇒ W_0=1.1eV ...

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Standard XII

Physics

The Photon Model

For intensity...

Question

For intensity I of a light of wavelength $5000 \circ A$ the photoelectron saturation current is 0.40 $μ A$ and stopping potential is 1.36 V, the work function of metal is

2.47 eV

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1.36 eV

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1.10 eV

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0.43 eV

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Solution

The correct option is C
1.10 eV

$B y u s i n g E = W_{0} + K_{m a x}$
$E = \frac{12375}{5000} = 2.475 e V a n d K_{m a x} = e V_{0} = 1.36 e V$
$S o 2.475 = W_{0} + 1.36 \Rightarrow W_{0} = 1.1 e V$

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For intensity I of a light of wavelength 5000∘A the photoelectron saturation current is 0.40 μ A and stopping potential is 1.36 V, the work function of metal is

The correct option is C 1.10 eV By using E=W0+Kmax E=123755000=2.475 eV and Kmax=eV0=1.36eV So 2.475=W0+1.36⇒W0=1.1eV

For intensity I of a light of wavelength $5000 \circ A$ the photoelectron saturation current is 0.40 $μ A$ and stopping potential is 1.36 V, the work function of metal is

The correct option is C
1.10 eV

$B y u s i n g E = W_{0} + K_{m a x}$
$E = \frac{12375}{5000} = 2.475 e V a n d K_{m a x} = e V_{0} = 1.36 e V$
$S o 2.475 = W_{0} + 1.36 \Rightarrow W_{0} = 1.1 e V$