Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

The uncertainty in the position of a moving bullet of mass $$10\ gm$$ is $$10^{-5}\ m$$. Calculate the uncertainty in its velocity

A
$$5.2 \times 10^{-25}\ m/s$$

B
$$3.0\times 10^{-28}\ m/s$$

C
$$5.2\times 10^{-28}\ m/s$$

D
$$3\times 10^{-28}\ m/s$$

Solution
Question 2
1 / -0

Imagine an atom made of a photon and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply Bohr atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength [Rydberg Constant = R] equal to

A
$$\frac{18}{5R}$$

B
$$\frac{5}{3R}$$

C
$$\frac{1}{3R}$$

D
$$\frac{2}{3R}$$

Solution
Question 3
1 / -0

For wave connected with proton, de-broglie wavelength change by 0.25% if its momentum change by $$ P_0 $$ initial momentum=

A
$$ 100 P_0 $$

B
$$ \dfrac {P_0}{400} $$

C
$$ 401 P_0 $$

D
$$ \dfrac {P_0}{100} $$

Solution

Using,
$$\lambda =\dfrac { h }{ p } $$ where $$\lambda =$$ wavelength
$$p=$$ Momentum
While Reducing wavelength by $$0.25$$% then momentum reduces by $$1$$ unit.
So, $$\left( 1-0.5\lambda \right) =\dfrac { h }{ p } $$
Initial Momentum $$=$$ $$401{ p }_{ 0 }$$
Question 4
1 / -0

Orbits of a particle moving in a circle are such that the perimeter of the orbit equals an integer number of de-Broglie wavelengths of the particle. For a charged particle moving in a plane perpendicular to a magnetic field, the radius of the nth orbital will therefore be proportional to

A
$$n^2$$

B
n

C
$$n^{1/2}$$

D
$$n^{1/4}$$

Solution
Question 5
1 / -0

Which of the following is an to a possible de-Broglie's wavelength of a particle, which moves inside a cubical box of side length $$L$$, without losing any energy (elasticity colliding with walls of cube)?

A

B

C

D
All of these

Solution
Question 6
1 / -0

If velocity of electron becomes equal to the velocity of light, then mass of electron will be

A
Decrease

B
Infinity

C
Remain same

D
Zero

Solution
Question 7
1 / -0

The wave length of a photon is twice the de Broglie wave length of the electron. If the speed of the electron is $$v_e = \dfrac{c}{{100}}$$. then the ratio of the energy of electron to that of photon $$\dfrac {E_e} {E_p}$$ is

A
$$10^{-1}$$

B
$$10^{-2}$$

C
$$10^{-3}$$

D
$$10^{-5}$$

Solution

$${\lambda _e} = \dfrac{h}{{{m_e}{v_e}}}$$
$$ = \dfrac{h}{{{m_e}\left( {c/100} \right)}}$$
$$ = \dfrac{{100}}{{{m_e}c}}$$
$$K.E,$$ $${E_e} = \dfrac{1}{2}{m_e}v_e^2$$
$${m_e}{v_e} = \sqrt {2{E_e}{m_e}} $$
$$\lambda = \dfrac{h}{{{m_e}{v^2}}}$$
$$ = \dfrac{h}{{2{m_e}{E_e}}}$$
$${E_e} = \dfrac{{{h^2}}}{{2{\lambda ^2}{m_e}}}$$
$$\dfrac{{{E_p}}}{{{E_e}}} = \dfrac{{hc}}{{2{\lambda _e}}} \times \dfrac{{2{\lambda ^2}{m_e}}}{{{h^2}}} = \dfrac{{{\lambda _e}{m_e}c}}{h}$$
$$ = \dfrac{{100h}}{{{m_e}c}} \times \dfrac{{{m_e}c}}{h} = 100$$
$$\dfrac{{{E_p}}}{{{E_e}}} = \dfrac{1}{{100}} = {10^{ - 2}}$$
For electron $${P_e} = {m_e}{v^2} = {m_e} \times \dfrac{c}{{100}} = {10^{ - 2}}$$
Hence,
option $$(B)$$ is correct answer.
Question 8
1 / -0

The de Broglie wavelength of a gas molecule at a temperature T K is:

A
$$\cfrac{h}{\sqrt{3mKT}}$$

B
$$\cfrac{h}{3mKT}$$

C
$$\cfrac{h}{\sqrt{2mKT}}$$

D
$$\sqrt{2mKT}$$

Solution
Question 9
1 / -0

The circumstance of first of hydrogen atom is s.Then the Broglie wavelength of electron to that orbit is

A
$$\dfrac {S} {2}$$

B
$$2S$$

C
$$S$$

D
$$3S$$

Solution

Broglie wavelength
$$\lambda = n\left( {2\pi R} \right)$$
$$\lambda = S$$
Hence,
option $$(c)$$ is correct answer.
Question 10
1 / -0

Wavenumber of radiations having frequency of $$ 4 \times 10^{14} Hz $$ will be:

A
$$ 1.33 \times 10^{6} \space cm^{-1} $$

B
$$ 1.33 \times 10^{4} \space cm^{-1} $$

C
$$ 9 \times 10^{-11} \space cm^{-1} $$

D
$$ 2.33 \times 10^{-6} \space cm^{-1} $$

Solution