KVPY 2020-21 SA Exam Previous Papers

Question : 37 of 80

Marks: +1, -0

A certain metal has a work function of Φ = 2 eV. It is irradiated first with 1 W of 400 nm light and later with 1 Wof 800 nm light. Among the following, the correct statement is:
[Given: Planck constant

(h) = 6.626 × 10^{–34} m^2\ kg\ s^{-1}

; Speed of light

(c) = 3 × 10^8 m\ s^{-1}]

Both colors of light give rise to same number of photoelectrons.
400 nm light gives rise to less energetic photoelectrons than 800 nm light.
400 nm light leads to more photoelectrons.
800 nm light leads to more photoelectrons.

Validate

Solution:

Energy associated with

1 {W}

400 {nm}

light

E ={h c}/{\λ}={6.626 \×10^{-34} \×3 \×10^{8}}/{400 \×10^{-9}} {J}

=3.1 {eV}

Likely, energy associated with

1 {W}

800 {nm}

light

{E}={{hc}}/{\λ}={6.626 \×10^{-34} \×3 \×10^{8}}/{800 \×10^{-9}}

=1.5 {eV}

Since, work function of the metal

=2 {eV}

Therefore, only

400 {nm}

light gives rise to ejection of photoelectrons.

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