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1

### JEE Main 2019 (Online) 12th January Morning Slot

MCQ (Single Correct Answer)
A particle of mass m moves in a circular orbit in a central potential field U(r) = $${1 \over 2}$$ kr2. If Bohr 's quantization conditions are applied, radii of possible orbitls and energy levels vary with quantum number n as :
A
rn $$\propto$$ $$\sqrt n$$, En $$\propto$$ n
B
rn $$\propto$$ $$\sqrt n$$, En $$\propto$$ $${1 \over n}$$
C
rn $$\propto$$ n, En $$\propto$$ n
D
rn $$\propto$$ n2, En $$\propto$$ $${1 \over {{n^2}}}$$

## Explanation

Force due to this field, F = $$- {{\partial U} \over {\partial r}}$$

F = $$- {\partial \over {\partial r}}\left( {{1 \over 2}k{r^2}} \right)$$ = -kr

For circular orbit, $${{m{v^2}} \over r}$$ = -kr

$$\Rightarrow$$ v $$\propto$$ r ..... (1)

Fron Bohr’s quantization condition

mvr = $${{nh} \over {2\pi }}$$ .....(2)

From (1) and (2),

$${r_n}$$ $$\propto$$ $${n^{{1 \over 2}}}$$

Given, U(r) = $${1 \over 2}$$kr2

$$\Rightarrow$$ En = $$- {1 \over 2}U\left( r \right)$$ = $$- {1 \over 4}k{r^2}$$

$$\Rightarrow$$ En $$\propto$$ n
2

### JEE Main 2019 (Online) 11th January Evening Slot

MCQ (Single Correct Answer)
In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is $$\lambda$$. If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be:
A
$${{25} \over {16}}$$ $$\lambda$$
B
$${{27} \over {20}}$$ $$\lambda$$
C
$${{16} \over {25}}$$ $$\lambda$$
D
$${{20} \over {27}}$$ $$\lambda$$

## Explanation

For M $$\to$$ L steel

$${1 \over \lambda }$$ = K $$\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right) = {{K \times 5} \over {36}}$$

for N $$\to$$ L

$${1 \over {\lambda '}}$$ = K$$\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right) = {{K \times 3} \over {16}}$$

$$\lambda ' = {{20} \over {27}}\lambda$$
3

### JEE Main 2019 (Online) 11th January Morning Slot

MCQ (Single Correct Answer)
A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980$$\mathop A\limits^ \circ$$. The radius of the atom in the excited state, in terms of Bohr radius a0 will be : (hc = 12500 eV$$\mathop A\limits^ \circ$$)
A
4a0
B
9a0
C
25a0
D
16a0

## Explanation

Energy of photon $$= {{12500} \over {980}} = 12.75eV$$

$$\therefore$$  Electron will excite to n = 4

Since 'R' $$\propto$$ n2

$$\therefore$$  Radius of atom will be 16a0
4

### JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)
Consider the nuclear fission

Ne20 $$\to$$ 2He4 + C12

Given that the binding energy/ nucleon of Ne20, He4 and C12 are, respectively, 8.03 MeV, 7.07 MeV and 7.86 MeV, identify the correct statement -
A
8.3 MeV energy will be released
B
energy of 11.9 MeV has to be supplied
C
energy of 12.4 MeV will be supplied
D
energy of 3.6 MeV will be released

## Explanation

Ne20      $$\to$$  2He4 + C12

Q – value, EB = (BE)react $$-$$ (BE)product

= (20 × 8.03) – ((2 × 7.07 × 4) + 7.86 × 12)

= 9.72 MeV

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