SELECT * FROM question_mgmt as q WHERE id=1192 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=2 AND subId=9 AND chapterId=44 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED) CBSE Free NCERT Solution of 11th chemistry Classification of Elements and Periodicity in Properties energy of an electron in the ground state of the h

Question:

Energy of an electron in the ground state of the hydrogen atom is –2.18 × 10–18 J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol–1.

Answer:

Ionization enthalpy is the minimum amount of energy which is needed to remove the loosely bound electron from a  isolated gaseous atom to form a cation.

It is given that the energy of an electron in the ground state of the hydrogen atom is –2.18 × 10–18 J.

Therefore, the energy required to remove that electron from the ground state of hydrogen atom is 2.18 × 10–18 J.

so Ionization enthalpy of atomic hydrogen = 2.18 × 10–18 J

Hence, ionization enthalpy of atomic hydrogen in terms of J mol–1 = 2.18 × 10–18 × 6.02 × 1023 J mol–1 = 1.31 × 106 J mol–1


SELECT ex_no,question,question_no,id,chapter FROM question_mgmt as q WHERE courseId='2' AND subId='9' AND ex_no!=0 AND status=1 and id!=1192 ORDER BY views desc, last_viewed_on desc limit 0,10
SELECT ex_no,question,question_no,id,chapter FROM question_mgmt as q WHERE courseId='2' AND subId='9' AND ex_no!=0 AND status=1 and id!=1192 ORDER BY last_viewed_on desc limit 0,10

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  • 4 months ago

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