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 $

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 - \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}} $ .