Question

Q. Show that energy E of a photon having wavelength λ can be written as E= (1240ev/ λ)nm.
Q. The energy needed to remove an electron from metallic sodium is 2.28ev. Does sodium show photo electric effect for red light having wavelength λ= 678 nm. What is cut off wavelength for photoelectric emission from sodium?

Answer 1

Answer on Question #58598-Physics-Quantum Mechanics

Q. Show that energy E of a photon having wavelength $\lambda$ can be written as $E = (1240\mathrm{ev} / \lambda)\mathrm{nm}$.

Solution

Show that energy E of a photon is

$$E = h\nu = h\left(\frac{c}{\lambda}\right).$$ $$\begin{array}{l} hc = 6.62606957 \cdot 10^{-34}J \cdot s \cdot 2.99792458 \cdot 10^{8} \frac{m}{s} = 1.9864457 \cdot 10^{-25}J \cdot m \\ = \left(\frac{1.9864457 \cdot 10^{-25}}{1.6 \cdot 10^{-19}}\right)eV \cdot m = 1.240 \cdot 10^{-5}eV \cdot m = 1240\,eV \cdot nm \\ \end{array}$$

Q. The energy needed to remove an electron from metallic sodium is 2.28ev. Does sodium show photoelectric effect for red light having wavelength $\lambda = 678$ nm. What is cut off wavelength for photoelectric emission from sodium?

Solution

$$E = h\nu = h\left(\frac{c}{\lambda}\right) = \frac{1240\,eV \cdot nm}{\lambda} = \frac{1240\,eV \cdot nm}{678\,nm} = 1.83\,eV.$$

It is less than 2.28ev, thus photoelectric effect will not occur.

The cut off wavelength for photoelectric emission from sodium is

$$\lambda_{cutoff} = \frac{1240\,eV \cdot nm}{2.28\,eV} = 543\,nm.$$

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