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======= ####32. Energy $E$ of a hydrogen atom with principal quantum number $n$ is given by $E = - 13.6 / n^{2} e V$ . The energy of photon ejected when the electron jumps from $n = 3$ state to $n = 2$ state of hydrogen is approximately

c3eec34ec6b1fad69db54a20ad4b2dca40c2aa54 (a) $1.9 e V$ .

(b) $1.5 e V$ .

(d) $3.4 e V$

[2004]

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Answer:

Correct Answer: 32. (a)

Solution:

(a) $Δ E = E_{3} - E_{2}$

$= \frac{- 13.6}{3^{2}} + \frac{13.6}{2^{2}}$

$= 13.6 \frac{1}{4} - \frac{1}{9} e V = 1.9 e V$

Atoms - Result Question 30

32. Energy $E$ of a hydrogen atom with principal quantum number $n$ is given by $E = - 13.6 / n^{2} e V$ . The energy of photon ejected when the electron jumps from $n = 3$ state to $n = 2$ state of hydrogen is approximately

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Atoms - Result Question 30

32. Energy E of a hydrogen atom with principal quantum number n is given by E=−13.6/n2eV. The energy of photon ejected when the electron jumps from n=3 state to n=2 state of hydrogen is approximately

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Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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32. Energy $E$ of a hydrogen atom with principal quantum number $n$ is given by $E = - 13.6 / n^{2} e V$ . The energy of photon ejected when the electron jumps from $n = 3$ state to $n = 2$ state of hydrogen is approximately