### JEE Mains Previous Years Questions with Solutions

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1

### AIEEE 2003

The orbital angular momentum for an electron revolving in an orbit is given by $\sqrt {l(l + 1)} {h \over {2\pi }}$. This momentum for an s-electron will be given by
A
zero
B
${h \over {2\pi }}$
C
$\sqrt 2 {h \over {2\pi }}$
D
$+ {1 \over 2}{h \over {2\pi }}$

## Explanation

For s-electron l = 0

$\therefore$ $\sqrt {l(l + 1)} {h \over {2\pi }}$

= $\sqrt {0(0 + 1)} {h \over {2\pi }}$

= $\sqrt 0 {h \over {2\pi }}$ = 0 (zero)
2

### AIEEE 2003

The number of d-electrons retained in Fe2+ (At no of Fe = 26) ion is
A
4
B
5
C
6
D
3

## Explanation

Fe (Z = 26) $\to$ e- = 26

Fe2+ (Z = 26) $\to$ e- = 24

Electronic configuration:

Fe (Z = 26) $\to$ 1s2 2s2 2p6 3s2 3p6 3d6 4s2

Fe2+ (Z = 26) $\to$ 1s2 2s2 2p6 3s2 3p6 3d6 4s0

as 4s is the outermost shell of Fe that's why e- will always removed first from 4s, so retained e- in d-shell is 6
3

### AIEEE 2002

In a hydrogen atom, if energy of an electron in ground state is -13.6 eV, then that in the 2nd excited state is
A
-1.51 eV
B
-3.4 eV
C
-6.04 eV
D
-13.6 eV

## Explanation

n = 1 means ground state

n = 2 means 1st excited state

n = 3 means 2nd excited state

Remember: nth excited state means n = n + 1

We know energy of an atom is given by

Energy (E) = -13.6 $\times$ ${{{Z^2}} \over {{n^2}}}$

= -13.6 $\times$ $1 \over 9$ [ $\because$ Z = atomic number = 1 for hydrogen atom]

= -1.51 ev
4

### AIEEE 2002

Uncertainty in position of a minute particle of mass 25 g in space is 10-5 m. What is the uncertainty in its velocity (in ms-1) (h = 6.6 $\times$ 10-34 Js)
A
2.4 $\times$ 10-34
B
0.5 $\times$ 10-34
C
2.1 $\times$ 10-28
D
0.5 $\times$ 10-23

## Explanation

According to Hysenberg's Uncertainty Principal $$\Delta x . \Delta p \ge { h \over {4\pi}}$$ Where $\Delta x$ is uncertainty in position and $\Delta p$ is uncertainty in momentum and as in momentum m (mass of object) is always same so we can say $$\Delta x . m. \Delta v \ge { h \over {4\pi}}$$ Mass of the particle (m) = 25 g = 0.025 kg

Position of the particle ($\Delta x$) = 10-5 m

$\Delta x . \Delta p$ = ${ h \over {4\pi}}$

$\Delta x . m . \Delta v$ = ${ h \over {4\pi}}$

$\Delta v$ = ${h \over {4 \times \Delta x \times m \times \pi }}$

$\Delta v$ = ${{6.6 \times {{10}^{ - 34}}} \over {4 \times {{10}^{ - 5}} \times 0.025 \times 3.14}}$

$\Delta v$ = 2.1 $\times$ 10-28