The photoelectric threshold wavelength for potassium (work function being 2 eV ) is: (Take h=6.6 × 10-34 J - s )

Question

The photoelectric threshold wavelength for potassium (work function being 2 eV ) is: (Take h=6.6 × 10-34 J - s ) (A) 310 nm (B) 620 nm (C) 1200 nm (D) 2100 nm

Tardigrade Admin · Accepted Answer

The minimum energy required for the emission of photoelectron from a metal is called the work function of that metal.

W = h v = \frac{h c}{λ}

Where

v

is frequency,

c

is speed of light,

λ

is wavelength.
Given,

W = 2 e V

= 2 \times 1.6 \times 1 0^{- 19} J

h = 6.6 \times 1 0^{- 34} J - s,

c = 3 \times 1 0^{8} m / s

\Rightarrow λ = \frac{h c}{W}

= \frac{6.6 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{8}}{2 \times 1.6 \times 1 0 ^{- 19}}

= 620 \times 1 0^{- 9} m = 620 nm

Note: If wavelength of incident light greater than

620 nm

no photoelectrons are emitted.

Q. The photoelectric threshold wavelength for potassium (work function being 2eV ) is : (Take h=6.6×10−34J−s )

310 nm

620 nm

1200 nm

2100 nm

Q. The photoelectric threshold wavelength for potassium (work function being $2 e V$ ) is :
(Take $h = 6.6 \times 1 0^{- 34} J - s$ )