Single Choice

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$$1.2 \ nm$$
Correct Answer
B$$3.215 \ nm$$
C$$7.28$$ $$\mathring{A}$$
D$$1.65$$ $$\mathring{A}$$

Solution

Kinetic energy= Quantum energy - Threshold energy

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

$$\dfrac{1}{2}mv^{2}= 1.6565\times 10^{-19}$$

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

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

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

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

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

Hence option $$A$$ is correct.

SIMILAR QUESTIONS

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