1

### AIPMT 2001

Energy released in nuclear fission is due to
A
some mass is converted into energy
B
total binding energy of fragments is more than the binding energy of parantal element
C
total binding energy of fragments is less than the binding energy of parental element
D
total binding energy of fragments is equal to the binding energy of parental element.

## Explanation

Energy released in nuclear fission is due to some mass is converted into energy.
2

### AIPMT 2001

The energy of hydrogen atom in nth orbit is En then the energy in nth orbit of singly ionised helium atom will be
A
4En
B
En/4
C
2En
D
En/2

## Explanation

As En = ${{ - 2{\pi ^2}m{K^2}{Z^2}{e^4}} \over {{n^2}{h^2}}}$

For helium Z = 2. Hence requisite answer is 4En.
3

### AIPMT 2001

Mn and Mp represent the mass of neutron and proton respectively. An element having mass M has N neutrons and Z protons, then the correct relation will be
A
M < {N $\cdot$ Mn + Z $\cdot$ Mp}
B
M > {N $\cdot$ Mn + Z $\cdot$ Mp}
C
M = {N $\cdot$ Mn + Z $\cdot$ Mp}
D
M = N {Mn + Mp}

## Explanation

Given : Mass of neutron = Mn
Mass of proton = Mp;
Atomic mass of the element = M ;
Number of neutrons in the element = N and number of protons in the element = Z.

We know that the atomic mass (M) of any stable nucleus is always less than the sum of the masses of the constituent particles.

Therefore, M $<$ [NMn + ZMp].

X is a neutrino, when b-particle is emitted.
4

### AIPMT 2001

The interplanar distance in a crystal is 2.8 $\times$ 10$-$8 m. The value of maximum wavelength which can be diffracted
A
2.8 $\times$ 10$-$8 m
B
5.6 $\times$ 10$-$8 m
C
1.4 $\times$ 10$-$8 m
D
7.6 $\times$ 10$-$8 m

## Explanation

2dsin$\theta$ = n$\lambda$

$\because$ $- 1 \le \sin \theta \le 1$

Therefore $\lambda$max = 2d

$\Rightarrow$ $\lambda$max = 2 $\times$ 2.8 $\times$ 10$-$8

= 5.6 $\times$ 10$-$8 m