Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

Suppose a   88226 Ra nucleus at rest and in ground state undergoes  α-decay to a   86222 Rn nucleus in its excited state. The kinetic energy of the emitted  α particle is found to be 4.44 MeV.   86222 Rn nucleus then goes to its ground state by  γ-decay. The energy of the emitted  γ photon is ____ keV.[Given : atomic mass of   88226 Ra = 226.005 u, atomic mass of   86222 Rn = 222.000 u, atomic mass of  α particle = 4.000 u, 1 u = 931 MeV/c2, c is speed of the light]

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

answer is 135.00.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

88226Ra⇒86222Rn+24HeΔm=mRa−mRn+ma=226.005−(222+4)=0.005uQ - Value =5×10−3×931MeV=4.655MeV, Since, KE∝1m⇒ K.E. of 86222Rn=4.44×4222=0.08MeV Energy of γ=(4.655−4.44−0.08)MeV=135keV
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring