QuestionJune 17, 2025

What is the change in energy of a hydrogen atom when it undergoes an electronic transition from n_(initial)=5 to n_(final)=2 2.09times 10^-18J 4.90times 10^-20J 4.57times 10^-19J 1.55times 10^-20J 1.55times 10^-19J

What is the change in energy of a hydrogen atom when it undergoes an electronic transition from n_(initial)=5 to n_(final)=2 2.09times 10^-18J 4.90times 10^-20J 4.57times 10^-19J 1.55times 10^-20J 1.55times 10^-19J
What is the change in energy of a hydrogen atom when it undergoes an electronic transition from
n_(initial)=5 to n_(final)=2
2.09times 10^-18J
4.90times 10^-20J
4.57times 10^-19J
1.55times 10^-20J
1.55times 10^-19J

Solution
4.4(262 votes)

Answer

2.53 \times 10^{-19} \, \text{J} (not listed among the options) Explanation 1. Use the energy level formula The energy of an electron in a hydrogen atom is given by E_n = -\frac{13.6 \, \text{eV}}{n^2}. Convert eV to Joules using 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}. 2. Calculate initial energy E_5 E_5 = -\frac{13.6}{5^2} \times 1.602 \times 10^{-19} \approx -8.72 \times 10^{-20} \, \text{J} 3. Calculate final energy E_2 E_2 = -\frac{13.6}{2^2} \times 1.602 \times 10^{-19} \approx -3.40 \times 10^{-19} \, \text{J} 4. Find change in energy \Delta E \Delta E = E_{final} - E_{initial} = (-3.40 \times 10^{-19}) - (-8.72 \times 10^{-20}) = -2.53 \times 10^{-19} \, \text{J}

Explanation

1. Use the energy level formula<br /> The energy of an electron in a hydrogen atom is given by $E_n = -\frac{13.6 \, \text{eV}}{n^2}$. Convert eV to Joules using $1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}$.<br />2. Calculate initial energy $E_5$<br /> $E_5 = -\frac{13.6}{5^2} \times 1.602 \times 10^{-19} \approx -8.72 \times 10^{-20} \, \text{J}$<br />3. Calculate final energy $E_2$<br /> $E_2 = -\frac{13.6}{2^2} \times 1.602 \times 10^{-19} \approx -3.40 \times 10^{-19} \, \text{J}$<br />4. Find change in energy $\Delta E$<br /> $\Delta E = E_{final} - E_{initial} = (-3.40 \times 10^{-19}) - (-8.72 \times 10^{-20}) = -2.53 \times 10^{-19} \, \text{J}$
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