Circular MotionHard
Question
The minimum frequency of photon required to break a particle of mass 15.348 amu into $4\alpha$ particles is $\_\_\_\_$ kHz .
[mass of He nucleus = 4.002 amu,
$1amu = 1.66 \times 10^{- 27}\text{ }kg,\text{ }h = 6.6 \times 10^{- 34}\text{ }J.s$ and $c = 3 \times 10^{8}\text{ }m/s$ ]
Options
A.$9 \times 10^{19}$
B.$9 \times 10^{20}$
C.$14.94 \times 10^{20}$
D.$14.94 \times 10^{19}$
Solution
$hv = (4 \times 4.002 - 15.348) \times 1.66 \times 10^{- 27} \times \left( 3 \times 10^{8} \right)^{2}v = 14.94 \times 10^{19}kHz$
Create a free account to view solution
View Solution FreeMore Circular Motion Questions
A bucket is whirled in a vertical circle with a string attached to it.The water in bucket does not fall down even when t...Which of the following quantities may remain constant during the motion of an object along a curved path :...The second’s hand of a watch has length 6 cm. Speed of end point and magnitude of difference of velocities at two perpen...A solid sphere is rotating about a diameter at an angular velocity ω. If it cools so that its radius reduces to 1/n...A spherically symmetric gravitational system of particles has a mass density ρ = where ρ0 is a constant. A tes...