Quantum Mechanics MCQ Quiz in मल्याळम - Objective Question with Answer for Quantum Mechanics - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Explanation:

 (given)

Case-1-Initial ground state energy without polarization

According to Pauli Exclusion principle, only two electrons filled in one state.

• Initial ground state energy 
• Now, 

Case-2-Final ground state energy after polarization

After polarization, only one electron filled in the state.

The change in ground state energy is 

So, the correct answer is .

ψ(x, t = 0) = 

 - 

 = 

 = 

Explanation:

We have 

Now > =1. Hence normalized.

Now we have to find the 2> =  +\frac{3}{7} + \frac{4}{21} \) .

Now using the relation that , we get:

=0+\frac{6}{7} \hbar^2 + \frac{8}{7} \hbar^2 = 2 \hbar^2 = 4\times\frac{\hbar^2}{2}\).

So the correct option is option 2).

Explanation:

WE are given a 2-dimensional box of length l.

There are electrons in it of mass m.

We have to find the ground state energy of 7 electrons.

So, we know that in 2-dimensional box energy eigen value is given by:

. 

For the ground state energy of 7 electrons the arrangement can be:

 .

hence the energy would be:

.

Simplifying it a bit we get:

.

Hence the correct option is option1).

Explanation:

 We have been given with and wavevector is . 

The total scattering cross-section for particle at l=0 states is given as :

. Putting all the values given in the above expression we get:

.

The correct option is option 3).

Explanation:

 The commutation and anti-commutation relations of Pauli matrices are given as:

 , where a subscript is for anti-commutation.

So, using the above relation we get:

 .

So, using the above relation we can write:

 .

Putting these in the final expression we get:

 .

The correct option is option 2).

Explanation:

We have two electrons hence the total spin will be:  .

Now,   .

Therefore .

Taking the expectation values on both the sides we get:

= \frac{1}{2} ( - - ) \).

We know that  = s_1(s_1 + 1) \hbar^2 , = s_2(s_2 + 1) \hbar^2, = s(s + 1) \hbar^2 \).

Also, we know that , as it is for electrons so for ground state the total spin will be s=0.

Hence putting all these in the above expression for expectation value we get:

= \frac{\hbar^2}{2} (0 - \frac{2}{2} (\frac{1}{2} +1) ) = - \frac{3 \hbar^2}{4} \).

The correct option is option 3).

Explanation:

At t=0 we have  = \frac{1}{\sqrt{2}} (|\psi_{+}> + |\psi_{-}> ) \). 

Now at some other time we will have the state evolved as  = \frac{1}{\sqrt{2}} (|\psi_{+}> e^{-\frac{i\epsilon t}{\hbar}}+ |\psi_{-}> e^{\frac{i\epsilon t}{\hbar}}) \).

If  we have  = \frac{1}{\sqrt{2}} (|\psi_{+}> e^{-\frac{i\pi}{3}}+ |\psi_{-}>e^{\frac{i\pi}{3}}) \).

Now in order to find the probability that this state will appear after the above stated time is;

|^2 =| \frac{e^{-i\pi/3} + e^{i\pi /3}}{2} |^2 = |\cos(\frac{\pi}{3})|^2 =\frac{1}{4} = 0.25 \).

We have used the property here that  =0 , =0 , = =1 \).

The correct option is option 2).

Explanation:

For  , we have columbic potential.

For this type of potential, we have  ,

which implies that  . .

Also, momentum transfer is given as .

 Hence, .

The correct option is option 2).

Concept:

The momentum operator is given by

p = - ih 

where h is the Plank constant.

Calculation:

e |x>

= [ ]|x>

= |x>

= |x> - a∇|x> +  (a∇)²|x> ... = |x-a>

X|x-a> = (x-a)|x-a>

The correct answer is an option (3).