The work function for a metal is 4.14 eV. The threshold wavelength for this metal surface is: $(h=6.62×10^{-34}Js, c=3×10^8 ms^{-1})$
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3000 Å
The threshold wavelength ($λ_{threshold}$) for a metal to it's work function ($\phi$) using the equations -
$\phi=\frac{hc}{λ_{threshold}}$
where,
$\phi$, work function = $4.14eV$ [given]
$∴λ_{threshold}=\frac{4.1357×10^{-15}×3×10^8}{4.14}$
$≃3000 Å$