What happens to ψ in a infinite potential well when the width is suddenly reduced to half its previous value ?
Will this instantly adjust ψ to the new size of the well or will it take some time to confine itself in this new well ? And is there a possibility of quantum tunneling here?
A parallel plate capacitor of circular cross section r>>d, and separation d. It is charged to potential V then disconnected from the circuit . What will be the work done in moving the capacitor from d to d1?
Answer:
Here the initial energy is 1/2CV^{2}
Where C=Aε/d
While moving the capacitor...
An atom can exist in two states , ground state of M and excited state of mass M+Δ , it goes to excited state by absorption of an photon, what is the photon frequency in the in the lab frame of reference ( where the atom is initially at rest )?
A simple energy equation gives the answer:
E=(...
Yes but U235 is stable and usual fission occurs only by bombarding with a neutron , thus making it U236 which is unstable.Also the question says it breaks up into two same products which means they should have the same number of nucleons. Therefore having an even number of nucleons is the only...
The binding energy (BE) per nucleon for 235U is 7.6 Mev. The 235U undergoes a nuclear fission to produce two fragments both having a BE of 8.5Mev. What is the energy released from a complete fission of 1kg of 235U (joules)?
Here I assumed that it breaks into two 118X element. Therefore...
Homework Statement
find the limit n\rightarrow∞ of 10n/ n!
Homework Equations
L hospital rule
The Attempt at a Solution
took log and separated the num and denom as:
n ln10-ln(n!)
n ln10-n ln(n)+n
1/n ( ln10 - ln(n)+1)
now i...
Ok I can just use the potentials to calculate the work done by :W=q.V
W = q.k[ -q/2a -q/2b + q/((2a)^2 + (2b)^2 )^0.5 ]
but should I multiply a negative sign to above equation because those were induced charges in moving q and were not present before , therefore there was negative work done.
Yes later I did take directions for the force and then for the upper right charge I resolved the force on the two axis with an unknown angle theta. Now that i have force as :
F= Kq^2[ (-1/(2x)^2 + cos/(2y)^2+(2x)^2 ) i + ( -1/(2y)^2+sin/(2y)^2+(2x)^2 ) j ]
then I did a line integral as for...