The minimum frequency of a photon required to | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The mass defect $\Delta m$ is the difference between the mass of the products and the initial mass.$\Delta m = (4 	imes 4.002 \ amu) - 15.348 \ a

Vedclass · Accepted Answer

$14.94 	imes 10^{20}$

Vedclass · Answer

(C) The mass defect $\Delta m$ is the difference between the mass of the products and the initial mass.$\Delta m = (4 	imes 4.002 \ amu) - 15.348 \ amu = 16.008 \ amu - 15.348 \ amu = 0.66 \ amu$.Convert the mass defect to $kg$: $\Delta m = 0.66 	imes 1.66 	imes 10^{-27} \ kg = 1.0956 	imes 10^{-27} \ kg$.The energy required $E = \Delta m c^2 = 1.0956 	imes 10^{-27} 	imes (3 	imes 10^8)^2 = 1.0956 	imes 10^{-27} 	imes 9 	imes 10^{16} = 9.8604 	imes 10^{-11} \ J$.Using $E = h
u$, the frequency $
u = E / h = (9.8604 	imes 10^{-11}) / (6.6 	imes 10^{-34}) \approx 1.494 	imes 10^{23} \ Hz$.Converting to $kHz$: $
u = 1.494 	imes 10^{23} / 10^3 = 1.494 	imes 10^{20} \ kHz$.

The minimum frequency of a photon required to break a particle of mass $15.348 \ amu$ into $4 \ \alpha$ particles is . . . . . . $kHz$.
[Mass of $He$ nucleus = $4.002 \ amu$, $1 \ amu = 1.66 \times 10^{-27} \ kg$, $h = 6.6 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$]

Similar Questions

If $200 \text{ MeV}$ of energy is released in the fission of one nucleus of ${ }_{92}^{236} U$, the number of nuclei that must undergo fission to release an energy of $1000 \text{ J}$ is

The example of nuclear fusion is

$A$ $\gamma$-ray photon creates an electron-positron pair. If the rest mass energy of an electron is $0.51 \, MeV$ and the total kinetic energy of the electron-positron pair is $0.78 \, MeV$,then the energy of the $\gamma$-ray photon in $MeV$ is:

Which one of the following is not correct?

An atomic power nuclear reactor can deliver $300 \, MW$. The energy released due to fission of each nucleus of uranium atom $U^{238}$ is $170 \, MeV$. The number of uranium atoms fissioned per hour will be (approximately):

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The minimum frequency of a photon required to break a particle of mass $15.348 \ amu$ into $4 \ \alpha$ particles is . . . . . . $kHz$. [Mass of $He$ nucleus = $4.002 \ amu$, $1 \ amu = 1.66 \times 10^{-27} \ kg$, $h = 6.6 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$]