A radio nuclide with half life 
second emits 
-particles of average kinetic energy 
11.25 eV . At an instant concentration of 
-particles at distance, 
from nuclide is 
per 
.
(i) Calculate number of nuclei in the nuclide at that instant.
(ii) If a small circular plate is placed at distance r from nuclide such that 
-particles strike the plate normally and come to rest, calculate pressure experienced by the plate due to collision of 
-particle. (Mass of 
-particle 
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
((i) 
, (ii) 
)
Let activity (rate of decay) of the nuclide be A nuclei per second. It means 
-particles are emitted per second. If a spherical surface of radius 
with center at position of nuclide be considered then 
-particle cross this surface per second. It means during an elemental time interval dt a number (A.dt) of 
-particle cross this surface. If velocity of 
-particles be 
then above calculated (A. dt) 
-particle are in a space having shape of a spherical shell of radius 
and radial thickness ( v dt ) as shown in figure.
Volume of this space 
(vdt)
∴ Concentration of 
-particles at distance 
from nuclide is
or activity of the nuclide, 
But activity, 
where N is number of nuclei
Hence, 
but decay constant 
Kinetic energy of 
-particle, 
substituting this value in equation (i),
(ii) At distance 
from the nuclide 
-particle cross unit area per second.
Let area of small circular plate be 
, then number of 
-particle striking the plate per second
Momentum of each particle just before collision is mv and after collision particles come to rest or momentum becomes zero.
∴ Momentum transferred to plate due to collision is
Due to transfer of momentum, the plate experiences a force which is equal to rate of transfer of momentum.
∴ Force, 
no. of particles striking per second
or
Pressure,
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