Calculating Recoil Energy and Velocity for Hydrogen Atom Transition [SOLVED]

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The discussion focuses on calculating the recoil energy and velocity of a hydrogen atom transitioning from the state n=4 to n=1 during photon emission. The energy difference between these states is calculated as 12.75 eV, leading to a recoil energy of 1.50 x 10^-10 J. A misunderstanding about the inclusion of rest energy in the recoil energy calculation was clarified, emphasizing that recoil energy pertains only to kinetic energy. The final resolution indicates that the initial approach was incorrect due to this misunderstanding. The problem was ultimately solved, confirming the correct interpretation of recoil energy.
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[SOLVED] recoil energy

Homework Statement



Calculate the recoil energy and velocity of a hydrogen atom in a transition from
the state « = 4 to the state AT = 1, in which a photon is emitted.

Homework Equations



E=-13.5/n^2

E=mc^2/sqrt(1-v^2/c^2)

The Attempt at a Solution



|E4-E1|=13.6-13.6/16 eV = 12.75 eV

Invoking the total energy,

Before emission
mc^2 = E recoil + |E4-E1|=> E recoil = 1.50 *10^-10 J

I couldn't get the correct answer. I wonder what is wrong? am I understanding the meaning of recoil energy wrongly?
 
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Recoil energy only pertains to the KE, and should not include the rest energy.
 
Thank you, I finally got it.
 
{\frac {-1/4\,i}{w}}
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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