Question

Verified by Toppr

Given, wavelength of light used, $λ=4950A˙$

Stopping potential, $$V_{o} =0.6V$$

According to Einstein potential equation we have;

$21 mv_{2}=eV_{o}=hv−V_{o}$........ (1)

ie, $eV_{o}=λhc −ϕ_{o}$

Where, $ϕ_{o}$ = work function

$V_{o}$ = stopping potential

For the first source

$λ_{1}=4950A˙=4950×10_{−10}$

$V_{o}=0.6v$

Putting these values in equation (1), we get

$1.6×10_{−19}×0.6=495×10_{−9}6.6×10_{−34×3×10_{8}} −ϕ_{o}$

$ϕ_{o}=3.04×10_{−19}J$

$=1.6×10_{−19}3.04×10_{−19} eV$

$=1.9eV$

Let $λ_{2}$ be the wavelength of the second source.

Stopping potential $V_{o}$ =1.1 v (given)

Therefore,

$1.6×10_{−19}×1.1=λ_{2}6.6×10_{−34×3×10_{8}} −3.04×10_{−19}J$

$1.76×10_{−19}=λ_{2}19.8×10_{−27} −3.04×10_{−19}$

$λ_{2}19.8×10_{−26} =4.8×10_{−19}$

$∴λ_{2}=4.8×10_{−19}19.8×10_{−26} =4.125×10_{−7}m$

$=4125A˙$

Solve any question of Dual Nature of Radiation And Matter with:-

0

0