Q31 / 160 · AP EAMCET 2010 · Chemistry

Which one of the following transitions of an electron in hydrogen atom emits radiation of the lowest wavelength?

  • A
    $n_2=\infty$ to $n_1=2$
  • B
    $n_2=4$ to $n_1=3$
  • Correct answer
    C
    $n_2=2$ to $n_1=1$
  • D
    $n_2=5$ to $n_1=3$

Solution

As the value of $n$ increases, the difference of energy between successive orbits decreases, ie, $E_2-E_1>E_3-E_2>E_4-E_3 \text { and so on... }$ $\because \quad \Delta E=\frac{h c}{\lambda}$ $\therefore \lambda$ is lowest for $n_2=2$ to $n_1=1$ transition.

Hence, the correct answer is option: 'C'.

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