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dexar [7]
3 years ago
5

Light that has a wavelength of 600 nm strikes a metal surface, and a stream of electrons is ejected from the surface. If light o

f wavelength 500 nm strikes the surface, the maximum kinetic energy of the electrons emitted from the surface will be
Smaller

The Same

Greater

Smaller by a factor of 5/6

Unmeasurable
Physics
1 answer:
Anni [7]3 years ago
8 0

Answer:

The maximum kinetic energy of the electrons emitted from the surface will be greater.

Option 3 is correct.

Explanation:

Given that,

Wavelength \lambda_{1}=600\ nm

Wavelength \lambda_{2}=500\ nm

We need to calculate the energy for first wavelength

Using formula of energy

E=hf

E=h\dfrac{c}{\lambda}

Put the value into the formula

E_{1}=6.67\times10^{-34}\times\dfrac{3\times10^{8}}{600\times10^{-9}}

E_{1}=3.33\times10^{-19}\ J

We need to calculate the energy for second wavelength

Put the value into the formula

E_{2}=6.67\times10^{-34}\times\dfrac{3\times10^{8}}{500\times10^{-9}}

E_{2}=4.002\times10^{-19}\ J

So, E₂>E₁

Hence, The maximum kinetic energy of the electrons emitted from the surface will be greater.

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