Question

The atomic masses of He and Ne are $4$ and $20$ a.m.u., respectively. The value of the de-Broglie wavelength of He gas at $-{73}^{\circ }C$ is "M" times that of the de-Broglie wavelength of $Ne$ at ${727}^{\circ }C.M$ is

Answer 1

Correct Answer: 5

Explanation:
Given:Atomic mass of $He=4a.m.u$Atomic mass of $Ne=20$ a.m. $U$${\lambda }_{He}$ at $-{73}^{\circ }C=M({\lambda }_{Ne}$ at ${727}^{\circ }C)$We have to find the value of $M$According to de-Broglie equation,$\lambda =\frac{h}{mv}$.....(i)where, $\lambda =$ de-Broglie wavelength$h=$ Planck's constant$m=$ Mass of the electron$v=$ Velocity of the electronAlso,$K.E=\frac{1}{2}m{v}^2$or$v=\sqrt{\frac{2KE}{m}}$.....(ii)On substituting the value of $v$ from equation (ii) to (i), we get.$\lambda =\frac{h}{m\times \sqrt{\frac{2KE}{m}}}$or,$\lambda =\frac{h}{\sqrt{2m\cdot KE}}$.....(iii)Also,Kinetic energy is directly proportional to temperature (T), i.e. K.E $\propto$ T or $K.E=xT$where, $x=$ Proportionality constantThus, equation (iii) can be written as$\lambda =\frac{h}{\sqrt{2m\times T}}$Now,Applying equation (iv) for He and Ne, we get${\lambda }_{He}=\frac{h}{\sqrt{2m_{He}\times T_{He}}}$.....(i${\lambda }_{Ne}=\frac{h}{\sqrt{2m_{Ne}\times T_{Ne}}}$.....(v)Dividing equation (iv) by (v), we get$\frac{{\lambda }_{He}}{{\lambda }_{Ne}}=\frac{\frac{h}{\sqrt{2m_{He}\times T_{He}}}}{\frac{h}{\sqrt{2m_{Ne}\times T_{Ne}}}}$or,$\frac{{\lambda }_{He}}{{\lambda }_{Ne}}=\sqrt{\frac{m_{Ne}{T}_{Ne}}{m_{He}{T}_{He}}}$...(vi)Now,Не$T_{\text{Не}}=-{73}^{\circ }C=-73+273K=200K$$T_{Ne}={727}^{\circ }C=727+273K=1000K$On putting the value in equation (vi), we get$\frac{{\lambda }_{He}}{{\lambda }_{Ne}}=\sqrt{\frac{20\times 1000}{4\times 200}}$or, $\frac{{\lambda }_{He}}{{\lambda }_{Ne}}=\sqrt{25}$or,${\lambda }_{He}=5\times {\lambda }_{Ne}=M\left({\lambda }_{Ne}\right)$Therefore, the value of M is 5.Hence, the correct answer is 5.00.