A perfectly reflecting mirror of mass M mounted on a spring constitutes a spring-mass system of angular frequency  $\Omega$  such that  $\frac{\text{4πMΩ}}{\text{h}}={10}^{24}{m}^{-2}$  with h as Planck’s constant. N photons of wavelength  $8\pi \mu m$ strike the mirror simultaneously at normal incidence such that the mirror gets displaced by  $\text{1μm}$ . If the value of N is  ${\text{x×10}}^{\text{12}}$, then the value of x is _________.
[consider the spring as massless]

If momentum of one photon is p and after reflection velocity gained by the mirror is v, by conservation of linear momentum we have
$$\begin{gathered} Np\widehat{i}=-Np\widehat{i}+mv\widehat{i} \\ \Rightarrow mv\widehat{i}=2pN\widehat{i} \\ \Rightarrow mv=2Np \\ \Rightarrow mv=2Np \end{gathered}$$ 
Since v is velocity of mirror attached to spring at mean position, it can be given as $v=A\Omega$
Where A is oscillation amplitude or maximum deflection of mirror from mean position and $\Omega$  is angular frequency of mirror spring system. Momentum of one incident photon is given as
 $$\begin{gathered} \Rightarrow p=\frac{h}{\lambda } \\ \Rightarrow mv=2Np \\ \Rightarrow mA\Omega =2N\frac{h}{\lambda } \\ \Rightarrow N=\frac{m\Omega }{h}\times \frac{\lambda A}{2} \end{gathered}$$
Given values are $\frac{m\Omega }{h}=\frac{{10}^{24}}{4\pi }{m}^{-2}$ and  $\lambda =8\pi \times {10}^{-6}m,$ so we have
 $N=\frac{{10}^{24}}{4\pi }\times \frac{8\pi \times {10}^{-6}\times {10}^{-6}}{2}$
 $\Rightarrow N={10}^{12}=x\times {10}^{12}$
 $\Rightarrow x=1$