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  • An electron in hydrogen atom in the second excited state jumps to the first and ground state of the atom. Find the ratio of the wavelengths emitted during this process.    </di

An electron in hydrogen atom in the second excited state jumps to the
first and ground state of the atom. Find the ratio of the wavelengths
emitted during this process.

 

 

 

 
 
 

Answers (1)

The determine the Ratio of wavelengths emitted we have
know the formula;-
\frac{1}{\lambda }= R\left [ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right ] 

 where, R=Rhydberg Constant
The value of n_{1}= 2 and n_{2}= 3, when in a hydrogen atom an electron in the second excited state jumps to the first. 
\frac{1}{\lambda _{1}}= R\left [ \frac{1}{2^{2}}- \frac{1}{3^{2}}\right ]
The wavelength of the electron when jumps to ground state;
\frac{1}{\lambda _{1}}= R\left [ \frac{1}{1^{2}}- \frac{1}{2^{2}}\right ]
Now,
the ratio of wavelengths emitted during this process is given as;
\frac{\frac{1}{\lambda _{1}}}{\frac{1}{\lambda_{2}}}= \frac{R\left [ \frac{1}{4}-\frac{1}{9} \right ]} {R\left [ \frac{1}{1} -\frac{1}{4}\right ]}
\frac{\lambda _{2}}{\lambda _{1}}= \frac{\left ( \frac{9-4}{36} \right )}{\left ( \frac{4-1}{4} \right )}
\frac{\lambda _{1}}{\lambda _{2}}= \frac{27}{5}

Posted by

Safeer PP

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