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Home  >>  JEEMAIN and NEET  >>  Chemistry  >>  Atomic Structure

The energy required for the ionisation of a certain atom is $3.44\times10^{-18}J$. The absorption of a photon of unknown wavelength ionises the atom and ejects an electron with velocity $1.03\times10^6$ m/sec. Calculate the wavelength of the incident radiation.

$(a)\;68.73nm\qquad(b)\;50.71nm\qquad(c)\;47nm\qquad(d)\;52.3nm$

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1 Answer

This is a case of photoelectric effect, work function being the ionization energy.
$\therefore$ Total Energy = KE + work function
$\large\frac{hc}{\lambda} = \large\frac{1}{2}mv^2+w$
$\large\frac{6.626\times10^{-34}\times3\times10^8}{\lambda} = \large\frac{1}{2}\times 9.1\times10^{-31}\times(1.03\times10^6)^2+3.44\times10^{-18}$
$\large\frac{2\times10^{-25}}{\lambda} = 3.92\times10^{-18}$
$\lambda = \large\frac{2\times10^{-25}}{3.92\times10^{-18}}$
=50.71nm
Hence the answer is (b)
answered Feb 5, 2014 by sharmaaparna1
 

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