Question

The photoelectric work function for a metal surface is
4.125 eV. The cut-off wavelength for this surface is
(a) 4125 Å (b) 3000 Å (c) 6000 Å (d) 2062 Å

Answer 1

Answer:

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Explanation:

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Answer 2

$\bigstar$ Answer $\bigstar$

$\checkmark$ Given:

⇒ Work Function ( W ) = 4.125 eV

$\checkmark$ To Find:

⇒ The cut-off wavelength ( λ₀ )

$\checkmark$ Solution:

We know that,

$\orange{\boxed{\boxed{\rm \lambda_{0}=\dfrac{hc}{W} }}}$

Here,

$\blue{\sf \implies h=Planck's \ constant}$

$\red{\sf \implies c=Velocity \ of \ light}$

$\blue{\implies \sf W=Work \ Function}$

$\red{\implies \sf \lambda_{0}=Threshold \ wavelength}$

NOTE:

Threshold wavelength and Cut-off wavelength are the same

According to the Question,

We are asked to find the cut-off wavelength for this surface

We know that,

$\orange{\implies \sf h=6.626 \times 10^{-34} \ Js}$

$\red{\sf \implies c=3 \times 10^{8} \ m/s}$

Given that,

$\blue{\sf \implies W = 4.125 \ eV}$

$\pink{\sf \implies W = 4.125 \times 1.6 \times 10^{-19} \ J}$

Since,

$\green{\longrightarrow \sf 1 \ eV=1.6 \times 10^{-19} \ J}$

Therefore,

Substituting the above values in the Formula

We get,

$\orange{\rm \implies \lambda_{0}=\dfrac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{4.125 \times 1.6 \times 10^{-19}} }$

On further Simplification,

We get,

$\red{\implies \rm \lambda_{0}=3000 \ A^{0}}$

$\orange{\boxed{\boxed{\rm \lambda_{0}=3000 \ A^{0}}}}$

Therefore,

The cut-off wavelength = 3000 Å

Hence,

$\pink{\large{\bold{\checkmark \ \rm Option \ (b) \ is \ correct}}}$