Ncert Exercises & Solutions

Consider a metal exposed to light of wavelength 600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm is used. Find the work function in eV.

Hint: Use Einstein's photoelectric equation.

Step 1: Find the work function of the metal.

Given,

For the first condition,

Wavelength of light, $λ = 600 n m$

And for the second condition,

Wavelength of light, $λ' = 400 n m$

Also, maximum kinetic energy for the second conditions is equal the twice fo the kinetic energy in first condition.

$\begin{aligned} i . e ., K_{max}^{'} & = 2 K_{max} \\ Here, K_{max} & = \frac{hc}{λ} - ϕ \\ \Rightarrow 2 K_{max} & = \frac{hc}{λ^{'}} - ϕ_{0} \\ \Rightarrow 2 (\frac{1230}{600} - ϕ) & = (\frac{1230}{400} - ϕ) [∴ hc ≃ 1240 eVnm] \\ \Rightarrow ϕ & = \frac{1230}{1200} = 1 .02 eV \end{aligned}$

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