The work function of a metallic surface is 5.01eV. Photoelectrons are emitted when light of wavelength 2000A∘ falls on it. The potential difference required to stop the fastest photoelectrons ish=4.14×10-15eVs

K.E=EK=hν-hν0=hν-ϕ0⇒EK=hcλ-ϕ0 ⇒EK=4.14×10-153×1082×10-7-5.01=6.21-5.01=1.2 eVSo, stopping potential required is 1.2V.

The work function of a metallic surface is 5.01eV. Photoelectrons are emitted when light of wavelength 2000A∘ falls on it. The potential difference required to stop the fastest photoelectrons ish=4.14×10-15eVs

K.E=EK=hν-hν0=hν-ϕ0⇒EK=hcλ-ϕ0 ⇒EK=4.14×10-153×1082×10-7-5.01=6.21-5.01=1.2 eVSo, stopping potential required is 1.2V.

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The work function of a metallic surface is $5.01 e V$ . Photoelectrons are emitted when light of wavelength $2000 \overset{\circ}{A}$ falls on it. The potential difference required to stop the fastest photoelectrons is
$(h = 4.14 \times 10^{- 15} e V s)$

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$1.2 V$

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answer is A.

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Detailed Solution

$K . E = E_{K} = h ν - h ν_{0} = h ν - ϕ_{0}$

$\Rightarrow E_{K} = \frac{h c}{λ} - ϕ_{0} \Rightarrow E_{K} = \frac{(4.14 \times 10^{- 15}) (3 \times 10^{8})}{2 \times 10^{- 7}} - 5.01 = 6.21 - 5.01 = 1.2 eV$

So, stopping potential required is $1.2 V$ .

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The work function of a metallic surface is 5.01eV. Photoelectrons are emitted when light of wavelength 2000A∘ falls on it. The potential difference required to stop the fastest photoelectrons ish=4.14×10-15eVs

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The work function of a metallic surface is $5.01 e V$ . Photoelectrons are emitted when light of wavelength $2000 \overset{\circ}{A}$ falls on it. The potential difference required to stop the fastest photoelectrons is
$(h = 4.14 \times 10^{- 15} e V s)$