Question

As we know that, ( lambda=frac{h}{p}=frac{h}{m V} )
( =frac{h}{sqrt{2 M K . E}} )
( Rightarrow lambda propto frac{1}{sqrt{M T}} quad[text { since } K . E . quad propto T] )
Hence( , ) for two gases,
( frac{lambda_{H e}}{lambda_{N e}}=sqrt{frac{M_{N e} T_{N e}}{M_{H e} T_{H e}}}=sqrt{frac{20 times 1000}{4 times 200}}=5 )
Hence( , frac{lambda_{mathrm{He}}}{lambda_{mathrm{Ne}}}=5 )
Hence the ratio of de-Broglie wavelength i.e. ( lambda_{mathrm{He}} ) ( : lambda_{mathrm{Ne}}=5: 1 )

# 3. The atomic masses of He and Ne are 4 and 20 a. mu, respectively. The value of de Broglie wavelength of He gas at -73°C is M times that of the Broglie wavelength of Ne at 727°C. M is (JEE 2013)

Solution