Threshold frequency, v_{0} is the minimum frequency which a photon must possess to eject an electron from a metal. It is different for different metals. When a photon of frequency 1.0\times 10^{15}\; s^{-1} was allowed to hit a metal surface, an electron having 1.988\times 10^{-19}J of kinetic energy was emitted. Calculate the threshold frequency of this metal. Show that an electron will not be emitted if a photon with a wavelength equal to 600 nm hits the metal surface.

Answers (1)

As per the quantum theory, hv=hv_{0}+\frac{1}{2}m\; v^{2}

Therefore, \frac{1}{2}m\; v^{2}=h(v-v_{0})

Where h represents the plank's constant, i.e. 6.626\times 10^{-34}J\; S

As per the information provided in the question, we must calculate v_{0} which represents the threshold frequency of photons.

v = frequency of incident photons. As per the information provided, v=1.0\times 10^{15}s^{-1}

\frac{1}{2}m\; v^{2}= electrons kinetic energy = 1.988\times 10^{-19}J

Therefore,

v_{0}=v-\left ( \frac{1}{2}m\; v^{2}/h \right )=1.0\times 10^{15}s^{-1}-\left ( 1.988\times 10^{-19}/6.626\times 10^{-34} \right )

= 1.0\times 10^{15} s^{-1} - 0.300030 \times 10^{15}

=6.9997\times 10^{14}s^{-1} (which is, in fact, the threshold frequency)

The relation between frequency and wavelength can be represented  by the equation,

v=c/\lambda ,, where λ represents the wavelength and c represents the speed of light, i.e. 3\times 10^{8}m/s

Here, in this case of the threshold frequency v_{0}=c/\lambda _{0}

Thus, the photon's maximum wavelength can be represented as,

\lambda _{0}=\frac{c}{v_{0}}=3\times 10^{8}m/s/6.9997\times 10^{14}s^{-1}

=4.36\times 10^{-7}m

Now, we know that 600\; nm=6\times 10^{-7}m which is certainly greater than the calculated value of the \lambda _{0}. Thus, the electron will not be emitted in case a photon which has a wavelength that is equal to 600 nm hits the surface of the metal.

 

 

 

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