Question

The work function of tungsten is $4.50eV$ . The wavelength of the fastest electron emitted when light whose photon energy is $5.50eV$ falls on the tungsten surface, is?
 $\left(A\right)12.27\stackrel{\circ }{A}$
 $\left(B\right)0.286\stackrel{\circ }{A}$
 $\left(C\right)1.24\stackrel{\circ }{A}$
 $\left(D\right)1.227\stackrel{\circ }{A}$

Answer 1

Hint: Here in this question we have to find the wavelength and for this first of all we have to find the kinetic energy, and for this the formula will be $K.E=E-\varphi$ . Then substituting this value, in the formula of wavelength which is given by $\lambda ={\displaystyle \frac{h}{\sqrt{2M_e\times K.E}}}$

Formula used:
Wavelength,
 $\lambda ={\displaystyle \frac{h}{\sqrt{2M_e\times K.E}}}$
 $\lambda$ , is the wavelength
 $h$ , is the Planck’s constant
 $M_e$ , is the mass of an electron
 $K.E$ , is the kinetic energy.

Complete step by step solution:
So we have the values given as
 $E=5.5eV$
 $\varphi =4.5eV$
Therefore, the kinetic energy will be equal to
 $\Rightarrow K.E=E-\varphi$
On substituting the values, we get
 $\Rightarrow K.E=\left(5.5-4.5\right)eV$
And on solving it, we will get the values as
 $\Rightarrow K.E=1eV$
Now we will solve for the wavelength,
So on substituting the values in the formula of wavelength, we get
 $\Rightarrow \lambda ={\displaystyle \frac{6.6\times {10}^{34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 1\times 1.16\times {10}^{-19}}}}$
And on solving the above equation, we get
 $\Rightarrow \lambda =1.24\times {10}^{9}m$
And it can also be written as
 $\Rightarrow 1.24\stackrel{\circ }{A}$
Therefore, the option $\left(C\right)$ is the correct answer.

Note:
Based on the principle of wave particle duality which states that every matter can behave as a wave as well as a particle. For if we consider it as a wave, it must be possessing some wavelength (property of wave) which is calculated from the concept of $\lambda ={\displaystyle \frac{h}{mv}}$ .