Question

To which orbit the electron will jump after absorbing 1.95*10^-18 J

Answer 1

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We know the formula that:

$E_n = -\frac{21.8x10^-^1^9}{n^2} J$

$and\frac{}{}\frac{}{}\frac{}{} -21.8x10^-^1^9 J\frac{}{} \frac{}{} \frac{}{} is\frac{}{} \frac{}{} \frac{}{} for\frac{}{} \frac{}{} H\frac{}{} \frac{}{} atom\frac{}{} \frac{}{} since\frac{}{} \frac{}{} n\frac{}{} \frac{}{} =1\frac{}{} \frac{}{}$

now, after absorbing 1.94 × 10⁻¹⁸ J =19.4 × 10⁻¹⁹ J of energy, the energy of the electron will be:

(-21.8 + 19.4) × 10⁻¹⁹ J = −2.4×10⁻¹⁹J

$\frac{-21.8x10^-^1^9}{n^2} = -2.4x10^-^1^9$

n² = 21.8/2.4 ≅ 9

so, n= 3