Dual Nature of Radiation and Matter Test

Result

Dual Nature of Radiation and Matter Test - 84

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The wavelength $$\lambda_e$$ an electron and $$\lambda_P$$ of a photon of same energy $$E$$ are related by:

A
$$\lambda_P\propto\displaystyle\dfrac{1}{1\sqrt{\lambda_e}}$$

B
$$\lambda_P\propto\lambda^2_e$$

C
$$\lambda_P\propto\lambda_e$$

D
$$\lambda_P\propto\sqrt{\lambda_e}$$

Solution
Question 2
1 / -0

A $$ \alpha $$ particle moves in a circular path of radius 0.83 cm in the presence of a magnetic field is $$ 0.25 Wb/m^2 $$. the de brogile wavelength association with the particle will be

A
$$ 10A^0 $$

B
$$ 0.01 A^0 $$

C
$$ 1 A^0 $$

D
$$ 0.1 A^0 $$

Solution

$$\begin{array}{l}R=\frac{m V}{q B}=\frac{P}{q B}\quad(P=m V \text { (momantum) }) \\R=\frac{P}{q B}\\0.83 \times 10^{-2}=\frac{P}{q \times 0.25} \\\text { (charge of } \alpha \text { particle =+2e charge) }\end{array}$$

$$\begin{aligned}p &=0.664 \times 10^{-21} \\\lambda=\frac{h}{P} &=\frac{6.6 \times 10^{-34}}{0.664 \times 10^{-21}} \\&=9.96 \times 10^{13}\\&=10\times 10^{-13}=0.01\mathrm{~A}^{0}\end{aligned}$$
Question 3
1 / -0

If an electron is moving with a velocity of $$1.1\times {10}^{6}m{s}^{-1}$$ de broglie wave length approximately is:

A
$$4.1\times {10}^{-11}m$$

B
$$4.9\times{10}^{-9}m$$

C
$$7\times{10}^{-10}m$$

D
$$3.306\times{10}^{-8}m$$

Solution
Question 4
1 / -0

The de Broglie wavelength is $$\lambda $$, the energy of electron is-

A
$$\frac{h}{\lambda} $$

B
$$\frac{h.c}{\lambda} $$

C
$$\frac{h^{2}}{2m\lambda^{2}} $$

D
None of these

Solution

$$\begin{array}{l}\lambda=\frac{h}{P}=\frac{h}{m V}\\\therefore V=\frac{h}{m \lambda}\\\therefore\text { Energy }=\frac{1}{2} m v^{2} \\=\frac{1}{2} \times m \times \frac{h^{2}}{m^{2}\lambda^{2}}\\.E=\frac{h^{2}}{2 m\lambda^{2}}\end{array}$$
Question 5
1 / -0

The de - Broglie wavelength of an electron accelerated by an electric potential of v volts is given by :

A
$$\lambda =\frac{1.23}{\sqrt{m}}$$

B
$$\lambda =\frac{1.23}{\sqrt{h}}m$$

C
$$\lambda =\frac{1.23}{\sqrt{V}}nm$$

D
$$\lambda =\frac{1.23}{V}$$

Solution
Question 6
1 / -0

According t Bohr's theory, the ratio n of the times taken by the electron in a hydrogen atom to complete one revolution in orbits corresponding to ground and first excited states is :

A
$$ 1 : 4 $$

B
$$ 4 : 1 $$

C
$$ 1 : 8 $$

D
$$ 8 : 1 $$

Solution
Question 7
1 / -0

The reaction, $$2B^-_{(aq)} + Sn^{2+}_{(aq)} \rightarrow Br_{2(1)} + Sn_{(s)}$$ with the standard potentials, $$E^o_{Sn} = -0.114 \,V, E^o_{Br_2} = +1.09 \,V$$, is

A
spontaneous in reverse direction

B
spontaneous in forward direction

C
at equilibrium

D
non-spontaneous in reverse direction

Solution
Question 8
1 / -0

The work function of the metal, if the kinetic energies of the photoelectrons are $$E_{1}$$ and $$E_{2}$$, with wavelengths of incident light $$\lambda_{1}$$ and $$\lambda_{2}$$, is

A
$$\dfrac {E_{1}\lambda_{1} - E_{2}\lambda_{2}}{\lambda_{2} - \lambda_{1}}$$

B
$$\dfrac {E_{1}E_{2}}{\lambda_{2} - \lambda_{1}}$$

C
$$\dfrac {(E_{1} - E_{2})\lambda_{1}\lambda_{2}}{(\lambda_{1} - \lambda_{2})}$$

D
$$\dfrac {\lambda_{1}\lambda_{2}E_{1}}{(\lambda_{1} - \lambda_{2})E_{2}}$$

Solution
Question 9
1 / -0

Two particles A and B of same mass have their de-Broglie wavelengths in the ratio $$X_A:X_B=K:1$$. Their potential energies $$U_A:U_B=1:K^2$$
The ratio of their total energies is $$E_A:E_B$$ is

A
$$K^2:1$$

B
$$1:K^2$$

C
$$K:1$$

D
$$1:K$$

E
$$1:1$$

Solution

According to question, $$\dfrac{X_A}{X_B}=\dfrac{K}{1}$$.......(i)
$$\dfrac{U_A}{U_B}=\dfrac{1}{K^2}$$
So, $$U_A=y$$ and $$U_B=K^2y$$
According to de-Broglie wavelength,
$$\lambda=\dfrac{h}{\sqrt{2mK_1}}$$..........(ii)
Here, $$K_1$$ is kinetic energy.
So, from Equations (i) and (ii),

$$\Rightarrow \dfrac{\left(\dfrac{h}{\sqrt{2mK_{1A}}}\right)}{\left(\dfrac{k}{\sqrt{2mK_{1B}}}\right)}=\dfrac{K}{1}\Rightarrow \sqrt{\dfrac{K_{1B}}{K_{1A}}}=\dfrac{K}{1}$$

$$\Rightarrow \dfrac{K_{1B}}{K_{1A}}=K^2$$
So, $$K_{1B}=K^2x$$ and $$K_{1A}=x$$

$$\because$$ Total energy, $$E=K_1+U$$
So, $$E_A=K_{1A}+U_A$$
and $$E_B=K_{1B}+U_B$$

or $$\dfrac{E_A}{E_B}=\dfrac{K_{1A}+U_A}{K_{1B}U_{1B}}=\dfrac{x+y}{K^2x+K^2y}$$

So $$\dfrac{E_A}{E_B}=\dfrac{1}{K^2}$$ or $$E_A:E_B=1:K^2$$
Question 10
1 / -0

Representing the stopping potential V along the y-axis and $$\dfrac{1}{\lambda}$$ along the x-axis for a given
photocathode, the curve is a straight line, the intercept on the y-axis is equal to

A
$$+W/e$$

B
$$-W/e$$

C
$$-We$$

D
$$e/W$$

Solution

As we know,
The maximum KE of the photoelectron is given by

$$\bigg(\dfrac{1}{2}mv^2\bigg)_{max}=hv-W$$

Now,
$$v=\dfrac{c}{\lambda}$$

and $$\bigg(\dfrac{1}{2}mv^2\bigg)=eV$$

$$eV=\dfrac{hc}{\lambda}-W$$

or $$V=\bigg(\dfrac{hc}{e}\bigg)\dfrac{1}{\lambda}-\dfrac{W}{e}$$

Since V is represented along the y-axis and $$(1/\lambda)$$ along the x-axis, the above equation represents a straight line.
The slope of straight line $$=\dfrac {hc}{e}$$

The intercept of straight-line $$=\dfrac{-W}{e}$$