Nuclear Structure

Sponsor Area

Question
ICSEENIPH12029763

Fill in the blank in the given nuclear reaction. 

____ + Al presubscript 23 presuperscript 27 space rightwards arrow space Mg presubscript 23 presuperscript 25 space plus space He presubscript 2 presuperscript 4

Solution

H presubscript 1 presuperscript 2 space plus space Al presubscript 23 presuperscript 27 space rightwards arrow space Mg presubscript 23 presuperscript 25 space plus space He presubscript 2 presuperscript 4

Sponsor Area

Question
ICSEENIPH12029811

Write a balanced nuclear reaction showing emission of a β- particle by 90Th234 (Symbol of daughter nucleus formed in the process is Pa.)

Solution

The balanced nuclear reaction is as given below:
         Th presubscript 90 presuperscript 234 space rightwards arrow space Pa presubscript 91 presuperscript 234 space plus space straight beta presubscript negative 1 end presubscript superscript straight o space plus space straight nu with bar on top

Question
ICSEENIPH12029938

When the cold junction of a certain thermo-couple was maintained at 20°C, its neutral temperature was found to be 180°C. Find its temperature of inversion.

Solution

Qnfraction numerator straight Q subscript 1 space plus space straight Q subscript 2 over denominator 2 end fraction

Here, 

Qn = 180o C ; Qc = 20o

180ofraction numerator straight Q subscript 1 space plus space 20 over denominator 2 end fraction  
Qi = 360 - 20 = 340o

Question
ICSEENIPH12030022

Alpha particles having kinetic energy of 1.8 MeV each are incident on a thin gold foil, form a large distance. Applying the principle of conservation of energy, find the closest distance of approach of the alpha particle from the gold nucleus. 

Solution

Kinetic energy of α-particle = Potential energy of repulsion between α-particle and nucleus.
straight K. straight E. space equals space fraction numerator 9 space straight x space 10 to the power of 9 space left parenthesis straight Z subscript straight e right parenthesis space 2 straight e over denominator straight r subscript straight e end fraction

1.8 space MeV equals space fraction numerator 9 space straight x space 10 to the power of 9 space left parenthesis Ze right parenthesis thin space 2 straight e over denominator straight r subscript straight o end fraction

straight r subscript straight o space equals space fraction numerator 9 space straight x space 10 to the power of 9 space straight x space 2 space straight x space Ze squared over denominator 1.8 space straight x space 10 to the power of 6 space straight x space 1.6 space straight x space 10 to the power of negative 19 end exponent end fraction

space space space equals space fraction numerator 9 space straight x space 10 to the power of 9 space straight x space 2 space straight x space 79 space straight x space left parenthesis space 1.6 space straight x space 10 to the power of negative 19 end exponent right parenthesis space squared over denominator 1.8 space straight x space 10 to the power of 6 space straight x space 1.6 space straight x space 10 to the power of negative 19 end exponent end fraction

space space space equals space fraction numerator 3640 space. space 32 space straight x space 10 to the power of 9 space straight x space 10 to the power of negative 36 end exponent over denominator 2.88 space straight x space 10 to the power of 6 space straight x space 10 to the power of negative 19 end exponent end fraction

   = 1264 x 10 -16

r0 = 1.264 x 10-13m

The distance of closest approach is 1.264 x 10-13 m.