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?
$\left( A \right)\,\,12.27\mathop A\limits^ \circ$
$\left( B \right)\,\,0.286\mathop A\limits^ \circ$
$\left( C \right)\,\,1.24\mathop A\limits^ \circ$
$\left( D \right)\,\,1.227\mathop A\limits^ \circ$

Answer 1

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 $K.E = E - \phi$ . Then substituting this value, in the formula of wavelength which is given by $\lambda = \dfrac{h}{{\sqrt {2{M_e} \times K.E} }}$

Formula used:
Wavelength,
$\lambda = \dfrac{h}{{\sqrt {2{M_e} \times K.E} }}$
$\lambda$ , 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
$E = 5.5eV$
$\phi = 4.5eV$
Therefore, the kinetic energy will be equal to
$\Rightarrow K.E = E - \phi$
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
$\Rightarrow \lambda = \dfrac{{6.6 \times {{10}^{34}}}}{{\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times 1 \times 1.16 \times {{10}^{ - 19}}} }}$
And on solving the above equation, we get
$\Rightarrow \lambda = 1.24 \times {10^9}m$
And it can also be written as
$\Rightarrow 1.24\mathop A\limits^ \circ$
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 $\lambda = \dfrac{h}{{mv}}$ .