If threshold wavelength (λ0) for ejection of electron from metal is 330 nm, then work function for the photoelectric emission is -

Question

If threshold wavelength (λ0) for ejection of electron from metal is 330 nm, then work function for the photoelectric emission is - (A) 1.2 × 1018 J (B) 1.2 × 10-20 J (C) 6 × 10-19 J (D) 6 × 10-12 J

Tardigrade Admin · Accepted Answer

W=h (C/λ0) ∴ W= (6.6 × 10-34×3×108/ 330×10-9) = (6.6 ×3×10-18/33) =0.6 ×10-18 =6×10-19 J

Q. If threshold wavelength $(λ^{0})$ for ejection of electron from metal is $330 nm$ , then work function for the photoelectric emission is -

$1.2 \times 1 0^{18} J$

$1.2 \times 1 0^{- 20} J$

$6 \times 1 0^{- 19} J$

$6 \times 1 0^{- 12} J$

$W = h \frac{C}{λ _{0}}$
$∴ W = \frac{6.6 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{8}}{330 \times 1 0 ^{- 9}}$
$= \frac{6.6 \times 3 \times 1 0 ^{- 18}}{33}$
$= 0.6 \times 1 0^{- 18}$
$= 6 \times 1 0^{- 19} J$

Q. If threshold wavelength (λ0) for ejection of electron from metal is 330nm, then work function for the photoelectric emission is -

1.2×1018J

1.2×10−20J

6×10−19J

6×10−12J