Work function of a metal is 3-0eV- It is illuminated by a light of wavelength 3 - 10-7m- Then the maximum energy of the electron is-

The correct option is C 1.13 eV Maximum energy =hcλ ϕ #160; (3× 10^ 7m=300nm) =1240300 3 (hc = #160; 1240 eV / X nm) =4.13 3 ...

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Standard XII

Physics

Huygens' Principle

Work function...

Question

Work function of a metal is 3.0eV. It is illuminated by a light of wavelength $3 \times 10^{- 7}$ m. Then the maximum energy of the electron is.

2.34 e V

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0.85 e V

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1.13 e V

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3.32 e V

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Solution

The correct option is C $1.13 e V$
$M a x i m u m e n e r g y = \frac{h c}{λ} - ϕ (3 \times 10^{- 7} m = 300 n m)$
$= \frac{1240}{300} - 3 (h c = 1240 e V / X n m)$
$= 4.13 - 3$
$= 1.13 e V .$
So, the answer is option (C).

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Work function of a metal is 3.0eV. It is illuminated by a light of wavelength 3×10−7m. Then the maximum energy of the electron is.

The correct option is C 1.13 eVMaximum energy=hcλ−ϕ(3×10−7m=300nm)=1240300−3 (hc=1240 eV/Xnm)=4.13−3=1.13eV.So, the answer is option (C).

Work function of a metal is 3.0eV. It is illuminated by a light of wavelength $3 \times 10^{- 7}$ m. Then the maximum energy of the electron is.

The correct option is C $1.13 e V$
$M a x i m u m e n e r g y = \frac{h c}{λ} - ϕ (3 \times 10^{- 7} m = 300 n m)$
$= \frac{1240}{300} - 3 (h c = 1240 e V / X n m)$
$= 4.13 - 3$
$= 1.13 e V .$
So, the answer is option (C).