Question

Photoelectrons are liberated by ultraviolet light of wavelength $3000\mathring{A}$ from a metallic surface for which the photoelectric threshold is $4000\mathring{A}$. The de-Broglie wavelength of electrons emitted with maximum kinetic energy is:

• A. $1.2nm$
• B. $3.215nm$
• C. $7.28$ $\mathring{A}$
• D. $1.65$ $\mathring{A}$

Answer 1

Answer: (A)

$1.2nm$

Kinetic energy= Quantum energy - Threshold energy

$K.E=\frac{6.625\times {10}^{-34}\times 3\times {10}^8}{3000\times {10}^{-8}}-\frac{6.625\times {10}^{-34}\times 3\times {10}^8}{4000\times {10}^{-8}}=1.6565\times {10}^{-19}$J

$\frac{1}{2}m{v}^2=1.6565\times {10}^{-19}$

$m^2\times v^2=2\times 1.6565\times {10}^{-19}\times 9.1\times {10}^{-31}$

$mv=5.49\times {10}^{-25}$

$\lambda =\frac{h}{mv}$

$\lambda =\frac{6.625\times {10}^{-34}}{5.49\times {10}^{-25}}=1.2\times {10}^{-9}$m

I.e 1.2 nm ($1nm={10}^{-9}m$)

Hence option $A$ is correct.