JEE Main 2014 · 11 Apr · Shift 1 · Chemistry Q40

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ChemistrySingle correct+4 / -1Question 40 of 90

Based on the equation :
$\Delta E=-2.0\times10^{-18}\text{ J}\left(\frac1{n_2^2}-\frac1{n_1^2}\right)$
the wavelength of the light that must be absorbed to excite hydrogen electron from level $n=1$ to level $n=2$ will be : ( $h=6.625\times10^{-34}$ Js, $C=3\times10^8$ ms $^{-1}$ )

1.325\times10^{-7}

1.325\times10^{-10}

2.650\times10^{-7}

5.300\times10^{-10}

Answer and worked solution

Official answer: A

$\Delta E=2.0\times10^{-18}\left(1-\frac14\right)=1.5\times10^{-18}$ J, so
$\lambda=\frac{hc}{\Delta E}=\frac{6.625\times10^{-34}\times3\times10^8}{1.5\times10^{-18}}=1.325\times10^{-7}\text{ m}.$

Topics:Bohr model Photoelectric effect and spectra

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JEE Main 2014 · 11 Apr · Shift 1, Chemistry question 40