When two monochromatic lights of frequency, ν and (ν)/2 are incident on a photoelectric metal, their stopping potential becomes V_S2 and V_s…

Question

When two monochromatic lights of frequency, ν and (ν)/2 are incident on a photoelectric metal, their stopping potential becomes V_S2 and V_s respectively. The threshold frequency for this metal is:

MARKS Content Team · Accepted Answer

Correct answer: (C) 3/2ν. Using Einstein's photoelectric equation we can write, For first case: hν=ϕ+eV_S2 ...1 For second case: h(ν)/2=ϕ+eV_S ...2 Using value of eV_S from equation(2) in equation(1), we can write, hν=ϕ+1/2(hν)/2-ϕ ⇒hν=ϕ+(hν)/4-(ϕ)/2 ⇒(ϕ)/2=(3hν)/4 ⇒ϕ=(3hν)/2 As we know, ϕ=hν_0 Therefore, ν_0=(3ν)/2

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

Solution

Relevant Questions

When two monochromatic lights of frequency, ν and ν2 are incident on a photoelectric metal, their stopping potential becomes VS2 and Vs respectively. The threshold frequency for this metal is:

Solution

Relevant Questions

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