Joint Entrance Examination

Graduate Aptitude Test in Engineering

1

MCQ (Single Correct Answer)

The half life period of radioactive element x is same as the mean life time of another radioactive element y. Initially they have the same number of atoms. Then :

A

x-will decay faster than y.

B

y-will decay faster than x.

C

x and y have same decay rate initially and later on different decay rate.

D

x and y decay at the same rate always.

Given, ($$\tau$$_{1/2})_{x} = ($$\tau$$)_{y}

Here, $$\tau$$_{1/2} = half-life period of radioactive element and $$\tau$$ = mean life period of radioactive element.

As we know the expression,

Half-life of the radioactive element x,

$${\tau _{1/2}} = {{\ln (2)} \over {{\lambda _x}}}$$

Mean life of the radioactive element y,

$$\tau = {1 \over {{\lambda _y}}}$$

Substituting the values in Eq. (i), we get

$${{\ln 2} \over {{\lambda _x}}} = {1 \over {{\lambda _y}}} \Rightarrow {\lambda _x} = 0.693{\lambda _y}$$

Initially they have same number of atoms,

$${N_x} = {N_y} = {N_0}$$

As we know that,

Activity, $$A = \lambda N$$

As, $${\lambda _x} < {\lambda _y} \Rightarrow {A_x} < {A_y}$$

Therefore, y will decay faster than x.

Here, $$\tau$$

As we know the expression,

Half-life of the radioactive element x,

$${\tau _{1/2}} = {{\ln (2)} \over {{\lambda _x}}}$$

Mean life of the radioactive element y,

$$\tau = {1 \over {{\lambda _y}}}$$

Substituting the values in Eq. (i), we get

$${{\ln 2} \over {{\lambda _x}}} = {1 \over {{\lambda _y}}} \Rightarrow {\lambda _x} = 0.693{\lambda _y}$$

Initially they have same number of atoms,

$${N_x} = {N_y} = {N_0}$$

As we know that,

Activity, $$A = \lambda N$$

As, $${\lambda _x} < {\lambda _y} \Rightarrow {A_x} < {A_y}$$

Therefore, y will decay faster than x.

2

MCQ (Single Correct Answer)

The temperature of an ideal gas in 3-dimensions is 300 K. The corresponding de-Broglie wavelength of the electron approximately at 300 K, is :

[m_{e} = mass of electron = 9 $$\times$$ 10^{$$-$$31} kg, h = Planck constant = 6.6 $$\times$$ 6.6 $$\times$$ 10^{$$-$$34} Js, k_{B} = Boltzmann constant = 1.38 $$\times$$ 10^{$$-$$23} JK^{$$-$$1}]

[m

A

6.26 nm

B

8.46 nm

C

2.26 nm

D

3.25 nm

Given, Planck's constant, h = 6.6 $$\times$$ 10^{$$-$$34} Js

Boltzmann constant, k_{B} = 1.38 $$\times$$ 10^{$$-$$23} J/K

Mass of an electron, m_{e} = 9 $$\times$$ 10^{$$-$$31} kg

Temperature of an ideal gas, T = 300 K

As we know that, de-Broglie wavelength,

$$\lambda = {h \over {mv}} = {h \over {\sqrt {2mE} }}$$ .... (i)

Here, E is the kinetic energy,

$$E = {{3{K_B}T} \over 2}$$

Substituting value of E in Eq. (i), we get

$$\lambda = {h \over {\sqrt {3m{K_B}T} }}$$

Substituting the given values in the above equation, we get

$$\lambda = {{6.6 \times {{10}^{ - 34}}} \over {\sqrt {3 \times 9 \times {{10}^{ - 31}} \times 1.38 \times {{10}^{ - 23}} \times 300} }}$$

= 6.26 nm

$$\therefore$$ The corresponding de-Broglie wavelength of an electron approximately at 300 K is 6.26 nm.

Boltzmann constant, k

Mass of an electron, m

Temperature of an ideal gas, T = 300 K

As we know that, de-Broglie wavelength,

$$\lambda = {h \over {mv}} = {h \over {\sqrt {2mE} }}$$ .... (i)

Here, E is the kinetic energy,

$$E = {{3{K_B}T} \over 2}$$

Substituting value of E in Eq. (i), we get

$$\lambda = {h \over {\sqrt {3m{K_B}T} }}$$

Substituting the given values in the above equation, we get

$$\lambda = {{6.6 \times {{10}^{ - 34}}} \over {\sqrt {3 \times 9 \times {{10}^{ - 31}} \times 1.38 \times {{10}^{ - 23}} \times 300} }}$$

= 6.26 nm

$$\therefore$$ The corresponding de-Broglie wavelength of an electron approximately at 300 K is 6.26 nm.

3

MCQ (Single Correct Answer)

Consider two separate ideal gases of electrons and protons having same number of particles. The temperature of both the gases are same. The ratio of the uncertainty in determining the position of an electron to that of a proton is proportional to :-

A

$${\left( {{{{m_p}} \over {{m_e}}}} \right)^{3/2}}$$

B

$$\sqrt {{{{m_e}} \over {{m_p}}}} $$

C

$$\sqrt {{{{m_p}} \over {{m_e}}}} $$

D

$${{{m_p}} \over {{m_e}}}$$

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

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

$$v = \sqrt {{{3KT} \over m}} $$

$${{\Delta {x_e}} \over {\Delta {x_p}}} = \sqrt {{{{m_p}} \over {{m_e}}}} $$

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

$$v = \sqrt {{{3KT} \over m}} $$

$${{\Delta {x_e}} \over {\Delta {x_p}}} = \sqrt {{{{m_p}} \over {{m_e}}}} $$

4

MCQ (Single Correct Answer)

A free electron of 2.6 eV energy collides with a H^{+} ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. (h = 6.6 $$\times$$ 10^{$$-$$34} Js)

A

1.45 $$\times$$ 10^{16} MHz

B

0.19 $$\times$$ 10^{15} MHz

C

1.45 $$\times$$ 10^{9} MHz

D

9.0 $$\times$$ 10^{27} MHz

For every large distance P.E. = 0

& total energy = 2.6 + 0 = 2.6 eV

Finally in first excited state of H atom total energy = $$-$$3.4 eV

Loss in total energy = 2.6 $$-$$ ($$-$$3.4) = 6 eV

It is emitted as photon

$$\lambda = {{1240} \over 6} = 206$$ nm

$$f = {{3 \times {{10}^8}} \over {206 \times {{10}^{ - 9}}}}$$ = 1.45 $$\times$$ 10^{15} Hz

= 1.45 $$\times$$ 10^{9} Hz

& total energy = 2.6 + 0 = 2.6 eV

Finally in first excited state of H atom total energy = $$-$$3.4 eV

Loss in total energy = 2.6 $$-$$ ($$-$$3.4) = 6 eV

It is emitted as photon

$$\lambda = {{1240} \over 6} = 206$$ nm

$$f = {{3 \times {{10}^8}} \over {206 \times {{10}^{ - 9}}}}$$ = 1.45 $$\times$$ 10

= 1.45 $$\times$$ 10

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