The photoelectric work function for a metal surface is 4.125 eV. The cut-off wavelength for this surface is

Question

The photoelectric work function for a metal surface is 4.125 eV. The cut-off wavelength for this surface is (A)  3000 mathringA  (B) 2062.5 mathringA  (C) 4125 mathringA  (D) 6000 mathringA

Tardigrade Admin · Accepted Answer

φ =hv= (hc/λ) ⇒ λ= (hc/φ)= (1242 eV.nm/4.125)≈ 3000 mathringA

Q. The photoelectric work function for a metal surface is $4.125 e V$ . The cut-off wavelength for this surface is

$3000 \overset{˚}{A}$

$2062.5 \overset{˚}{A}$

$4125 \overset{˚}{A}$

$6000 \overset{˚}{A}$

$ϕ = h v = \frac{h c}{λ}$
$\Rightarrow λ = \frac{h c}{ϕ} = \frac{1242 e V . nm}{4.125} \approx 3000 \overset{˚}{A}$

Q. The photoelectric work function for a metal surface is 4.125eV. The cut-off wavelength for this surface is

3000A˚

2062.5A˚

4125A˚

6000A˚