MCQ
Consider a hydrogen atom with its electron in the $n ^{\text {th }}$ orbital. An electromagnetic radiation of wavelength $90 \ nm$ is used to ionize the atom. If the kinetic energy of the ejected electron is $10.4 \ eV$, then the value of $n$ is ( $hc =1242 \ eV \ nm$ )
  • A
    $1$
  • $2$
  • C
    $3$
  • D
    $4$

Answer

Correct option: B.
$2$
b
$E _{\text {pboton }}= E _{\text {ionize arom }}+ E _{\text {linetic energy }}$

$\frac{1242}{90}=\frac{13.6}{ n ^2}+10.4$

from this, $n =2$

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