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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Atoms

Electrons of energies 12.09 eV can cause radiation to be emitted from hydrogen atom. calculate the principal quantum number of the orbit to which the electron in the hydrogen atom is raised and the wavelength of radiation emitted if it drops back to the ground state .

$(a)\;2 \times 1026 A^{0}\qquad(b)\;2 \times 2052 A^{0}\qquad(c)\;3\; and \;1026 A^{0}\qquad(d)\;3 \times 2052 A ^{0}$

1 Answer

Answer : (c) $\;3\;and\;1026 A^{0}$
Explanation :
$\varepsilon_{n}^{H}=-\large\frac{13.6}{n^2}\;eV$
$\varepsilon_{1}^{H}=-\large\frac{13.6}{1^2}\;eV=-13.6\;eV \quad \; (n=1)$
$\varepsilon_{2}^{H}=-\large\frac{13.6}{2^2}\;eV=-3.4\;eV \quad \; (n=2)$
$\varepsilon_{3}^{H}=-\large\frac{13.6}{3^2}\;eV=-1.51\;eV \quad \; (n=3)$
Now , $\;\varepsilon_{2}^{H}-\varepsilon _{1}^{H}=-3.4\;eV+13.6eV=10.2 eV$
Now , $\;\varepsilon_{3}^{H}-\varepsilon _{1}^{H}=-1.51\;eV+13.6eV=12.09 eV$
Hence electrons will be raised to the table of principal quantum number 2 and 3 corresponding to radiation energies 10.2eV and 12.09eV respectively . wavelength of radiation emitted for
$\lambda=\large\frac{6.62\times10^{-34}\times3\times10^{8}}{12.09\times1.6\times10^{-19}}=1026 A^{0}\;.$
answered Feb 22, 2014 by yamini.v
 

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