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Two hydrogen atoms collide head on and end up with zero kinetic energy.Each atom then emits a photon of wavelength 121.6nm.Which transition leads to this wavelength ?How fast were the hydrogen atoms traveling before collision?$R_H=1.097\times 10^7m^{-1}$ and $m_H=1.67\times 10^{-27}Kg$

$\begin{array}{1 1}(a)\;4.43\times 10^4m/s&(b)\;2.43\times 10^4m/s\\(c)\;4.43\times 10^2m/s&(d)\;1.6\times 10^4m/s\end{array}$

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1 Answer

$\large\frac{1}{\lambda}$$=R_H\big[\large\frac{1}{1^2}-\frac{1}{h^2}\big]$
$\large\frac{1}{121.6\times 10^{-9}}=$$1.097\times 10^7\big[\large\frac{1}{1^2}-\frac{1}{n^2}\big]$
$n=2$
The energy released is due to collision and all the kinetic energy is released in form of photon.
$\large\frac{1}{2}$$mu^2=\large\frac{hc}{\lambda}$
$\large\frac{1}{2}$$1.67\times 10^{-27}\times u^2=\large\frac{6.626\times 10^{-34}\times 3\times 10^8}{121.6\times 10^{-9}}$
$u=4.43\times 10^4m/s$
Hence (a) is the correct answer.
answered Jan 16, 2014 by sreemathi.v
 

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