KVPY 2019 SB SX Exam Previous Papers

Question : 95 of 120

Marks: +1, -0

Four electrons, each of mass m

_e

are in a one dimensional box of size L. Assume that the electrons areinteracting, obey the Pauli exclusion principle and are described by standing de Broglie waves confined within the box. Define

α = h^2\/8m_eL ^2

and U

_0

to be the ground state energy. Then -

The energy of the highest occupied state is 16α
$U_0 = 30α$
the total energy of the first excited state is $U_0 + 9α$
The total energy of the second excited state is $U_0 + 8α$

Validate

Solution:

For standing wave

n\ λ/2=L

P = h/λ=nh/{2L}

So Energy

F = P^2/{2m_e}={n^2h^2}/{8m_eL^2}=n^2α

For ground state

E = α = u_0

For first excited state

E = 2^2α = 4u_0

u_0 + 3α

For second excited sate

E = 3^2 α = 9α

= u_0 + 8α

Go to Question:

Prev Question

Next Question