Energy nth orbit, -k/(n)2 ;k:constnt, then as n increases how can the | energy | increase?
Asked by anuragchakraborty | 28th Feb, 2010, 10:07: PM
The enrgy of the n = (infinity) level is the highest since it has the highest potential energy.
This is taken as E = 0
As we come closer towards the nucleus, the enrgy decreases and hence becomes negative.
The negative sign is of significance here.
Energy nth orbit, -k/(n)2 ;k:constnt, t
Since as n increases the deniminatotr will increase. So, the fraction will decraese.
But since the negative sign is there, it menas that E is increasing
For example: -3< -2
Answered by | 2nd Mar, 2010, 10:13: AM
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