When two monochromatic lights of frequency, ν and ν 2 are incident on a photoelectric metal, their stopping potential becomes V S 2 and V s respectively. The threshold frequency for this metal is:

Question

Aicharya · Accepted Answer

Using Einstein's photoelectric equation we can write,

For first case:

$h ν = ϕ + e (\frac{V_{S}}{2}) . . . (1)$

For second case:

$h \frac{ν}{2} = ϕ + e (V_{S}) . . . (2)$

Using value of $e (V_{S})$ from equation(2) in equation(1), we can write,

$h ν = ϕ + \frac{1}{2} (\frac{h ν}{2} - ϕ)$
$\Rightarrow h ν = ϕ + \frac{h ν}{4} - \frac{ϕ}{2}$
$\Rightarrow \frac{ϕ}{2} = \frac{3 h ν}{4}$
$\Rightarrow ϕ = \frac{3 h ν}{2}$

As we know, $ϕ = h ν_{0}$

Therefore, $ν_{0} = \frac{3 ν}{2}$

When two monochromatic lights of frequency, ν and ν 2 are incident on a photoelectric metal, their stopping…

Solution