# If the work function is 6 eV, then the maximum wavelength of the incident radiation to remove electrons from that metal will be: (h = 6.624 × 10-34 Js

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If the work function is 6 eV, then the maximum wavelength of the incident radiation to remove electrons from that metal will be: (h = 6.624 × 10-34 Js, c = 3 × 108 m/sec)
1. 3145 Å
2. 2500 Å
3. 2071 Å
4. 1895 Å

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Correct Answer - Option 3 : 2071 Å

Concept:

Photoelectric Effect:

• The phenomena by which rays of light falling on the metal surface make electrons ejected out of the metal surface is known as the Photoelectric Effect.
• Work Function (ϕ) =The minimum energy required to eject an electron out of metal is known as work function.
• Threshold Frequency: The minimum frequency required so that the photoelectric effect can took place is called threshold frequency.
• It is given as

ϕ = h ν0

where ν0 is the threshold frquency,  the minimum frequency required for a light ray to eject an electron from that metal, h is Planck's constant.

This can also be written as

$\phi = \frac{hc}{λ }$

λ is the threshold wavelength or the maximum wavelength of the light so that the photoelectric effect can happen,  c is the speed of light.

Calculation:

Given

Work function  ϕ = 6 eV

Plancks constant h = 6.624 × 10-34 Js

c = 3 × 108 m / sec

$6 \times 1.602 × 10 ^{-19} J= \frac{(6.624 × 10 ^{-34})(3 × 10 ^8 )}{λ }$

(1 ev = 1. 602 × 10 -19 J )

$\implies λ = \frac{(6.624 × 10 ^{-34})(3 × 10 ^8 )}{6 \times 1.602 × 10 ^{-19} J }$

λ ≈ 2071 Å

(1 Å = 10 -10 m )

So, the correct option is 2071 Å.

The  Einstein equation for the photoelectric effect is given as:

K = hν - ϕ

Where h is Planck constant, ν is the frequency of light falling K is the kinetic energy of an electron ejected.

The light is said to be made up of small packets of energy each having energy equal to h ν, where ν is the frequency of the source light.

• The Equation can also be written as

K = hc/ λ - ϕ

c is the speed of light, c = 3 × 108 m/s, λ is the wavelength.