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Physics-

General

Easy

Question

If the series limit of Lymen series for Hydrogen atom is equal to the series limit of Balmer series for a hydrogen like atom, then atomic number o this hydrogen like atom will be

1
2
3
4

The correct answer is: 2

By using $fraction numerator 1 over denominator lambda end fraction equals R Z to the power of 2 end exponent open square brackets fraction numerator 1 over denominator n subscript 1 end subscript superscript 2 end superscript end fraction minus fraction numerator 1 over denominator n subscript 2 end subscript superscript 2 end superscript end fraction close square brackets$
For Hydrogen atom $fraction numerator 1 over denominator left parenthesis lambda H subscript m i n end subscript open square brackets fraction numerator 1 over denominator 1 to the power of 2 end exponent end fraction minus fraction numerator 1 over denominator infinity end fraction close square brackets end fraction$
.....(i)
For hydrogen like atom $open parentheses fraction numerator 1 over denominator lambda subscript m i n end subscript end fraction left parenthesis right parenthesis subscript a t o m end subscript to the power of 2 end exponent open parentheses fraction numerator 1 over denominator 2 to the power of 2 end exponent end fraction minus fraction numerator 1 over denominator infinity end fraction close parentheses close parentheses$
$rightwards double arrow left parenthesis lambda a t o m fraction numerator 4 over denominator R Z to the power of 2 end exponent end fraction subscript m i n end subscript blank$ ....(ii)
From equation (i) and (ii) $fraction numerator 1 over denominator R end fraction equals fraction numerator 4 over denominator R Z to the power of 2 end exponent end fraction rightwards double arrow Z equals 2.$

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On bombarding $U to the power of 235 end exponent$ by slow neutrons, 200MeV energy is released. If the power output of atomic reactor is 1.6 MW, then the rate of fission will be

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