Question

Maximum kinetic energy of emitted photoelectrons from a metal is 0.9ev and work function is 2.2ev then the wave length of incident radition is

Answer 1

Concept Introduction: Light has two types of properties wave and particle.

Given:

We have been Given: Work Function is

$2.2eV$

Kinetic Energy is

$0.9eV$

To Find:

We have been to Find: the wave length of incident radiation is.

Solution:

According to the problem,

$ke = hv - h {v}^{°} \\ 0.9 = hv - 2.2 \\ hv = 2.2 + 0.9 \\ hv = 3.1$

now, we know that,

$energy = hv = \frac{hc}{lambda}$

therefore,

$\frac{hc}{lambda} = 3.1 \\ \frac{6.6 \times 3 \times {10}^{ - 34} \times {10}^{8} }{lambda} = 3.1 \\ lambda = \frac{ 61.38 \times {10}^{ - 26} }{1.6 \times {10}^{ - 19} } \\ lambda = 38.36 \times {10}^{ - 7} m$

Final Answer: The wavelength is

$383.6nm$

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