Normal 0false false falseEN-US X-NONE X-NONE/* Style Definitions */table.MsoNormalTable{mso-style-name:Table Normal;mso-tstyle-rowband-size:0;mso-tstyle-colband-size:0;mso-style-noshow:yes;mso-style-priority:99;mso-style-parent:;mso-padding-alt:0in 5.4pt 0in 5.4pt;mso-para-margin-top:0in;mso-para-margin-right:0in;mso-para-margin-bottom:10.0pt;mso-para-margin-left:0in;line-height:115%;mso-pagination:widow-orphan;font-size:11.0pt;font-family:Calibri,sans-serif;mso-ascii-font-family:Calibri;mso-ascii-theme-font:minor-latin;mso-hansi-font-family:Calibri;mso-hansi-theme-font:minor-latin;mso-bidi-font-family:Times New Roman;mso-bidi-theme-font:minor-bidi;}When a hydrogen atom is raised from the ground state to an excited state,(a) P.E decreases and K.E. increases(b)P.E. increases and K.E decreases(c) Both K.E. and P.E. decrease(d) Absorption spectrum
Amit Saxena , 11 Years ago
Grade upto college level
2 Answers
Hrishant Goswami
Last Activity: 11 Years ago
(b)
, where, r is the radius of orbit which increases as we move from ground to an excited state. Therefore, when a hydrogen atom is raised from the ground state, it increases the value of r. As a result of this, P.E. increases (decreases in negative) and K.E. decreases.
Kushagra Madhukar
Last Activity: 4 Years ago
Dear student,
Please find the solution to your problem.
PE = – KZe2/r
KE = KZe2/2r
in excited state, r increases.
Hence, PE increases and KE decreases.
Thanks and regards,
Kushagra
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