Dual Nature of Radiation and Matter Test

Result

Dual Nature of Radiation and Matter Test - 45

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 slope of graph drawn between stopping potential and frequency of incident light for a given surface will be:-

A
h

B
h/e

C
eh

D
e

Solution

$$eV_{o}= h V [V_{o}= Stooping\ potential, V= frequency ]$$
$$\dfrac{V_{o}}{V}= \dfrac{h}{e} = $$ side
Question 2
1 / -0

de-Broglie wavelength associated with an electron revolving in the $$n^{th}$$ state of hydrogen atom is directly proportional to

A
$$n$$

B
$$\cfrac{1}{n}$$

C
$$n^2$$

D
$$\cfrac{1}{n^2}$$

Solution

$$ V_n = 2.165 \times 10^6(\cfrac{z}{n})$$
$$ V_n \propto \cfrac{1}{n}$$
$$ mV_n \propto \cfrac{1}{n}$$
$$\therefore \lambda = \cfrac{h}{mV_n} \propto n$$
Question 3
1 / -0

The de-Brogile wavelength corresponding to the root mean square velocity of the hydrogen molecule at $$20^{o}C$$.

A
$$2.04\ \mathring {A}$$

B
$$1.04\ \mathring {A}$$

C
$$3.04\ \mathring {A}$$

D
$$4.04\ \mathring {A}$$

Solution

Given that,

$$T=20+273=293\,K$$

Molecular mass of $${{H}_{2}}=2\times 1.00794=2.01588\times 1.66\times {{10}^{-27}}\,kg$$

We know that,

$$ \dfrac{1}{2}mv_{rms}^{2}=\dfrac{3}{2}kT $$

$$ {{v}_{rms}}=\sqrt{\dfrac{3kt}{m}} $$

Now, the de Broglie wave length

$$ \lambda =\dfrac{h}{m{{v}_{rms}}} $$

$$ \lambda =\dfrac{h}{\sqrt{3mkT}} $$

$$ \lambda =\dfrac{6.63\times {{10}^{-34}}}{\sqrt{3\times 2.01588\times 1.66\times {{10}^{-27}}\times 1.38\times {{10}^{-23}}\times 293}} $$

$$ \lambda =\dfrac{6.63\times {{10}^{-34}}}{\sqrt{4059.20\times {{10}^{-50}}}} $$

$$ \lambda =\dfrac{6.63\times {{10}^{-34}}}{63.712\times {{10}^{-25}}} $$

$$ \lambda =0.104\times {{10}^{-9}} $$

$$ \lambda =1.04\overset{\circ }{\mathop{A}}\, $$

Hence, the de Broglie wave length is $$1.04\overset{\circ }{\mathop{A}}\,$$
Question 4
1 / -0

The equation $$E=pc$$ is valid

A
For an electron as well as for a photon

B
For an electron but not for a photon

C
For an photon but not for an electron

D
Neither for an electron nor for a photon
Question 5
1 / -0

Electron has energy of 100 eV what will be its wavelength

A
$$1.2 \mathring { A }$$

B
$$10 \mathring { A }$$

C
$$100 \mathring { A }$$

D
$$1 \mathring { A }$$

Solution

Energy $$E=100ev=100\times 1.6 \times 10^{-19} Joule$$
Wavelength, $$\lambda = \dfrac{h}{\sqrt[2]{2mE}}=\dfrac{6.62\times 10^{-34}Js}{\sqrt[2]{2\times 9.1\times 10^{-31} Kg \times 1.6\times 10^{-17}J}}=1.226\times 10^{-10} m=1.2Angstrom$$
Option A is correct.
Question 6
1 / -0

Photoelectric there should wavelength of photo sensitive material is 3700 A, the work function in eV is $$(h=6.63 \times 10^{-34} Js)$$ :

A
1.55 eV

B
3.35 eV

C
4.85 eV

D
5.25 eV

Solution

$$\displaystyle {{hc} \over \alpha } = w$$ function-

$$ = \displaystyle {{6.64 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {3700 \times {{10}^{ - 10}} \times 1.6 \times {{10}^{ - 19}}}}$$

$$ = 3.36ev$$
Question 7
1 / -0

What is de-Broglie wavelength of electron having energy $$10\ keV$$?

A
$$0.12$$ $$A^{\circ}$$

B
$$1.2$$ $$A^{\circ}$$

C
$$12.2$$ $$A^{\circ}$$

D
none of these.

Solution
Question 8
1 / -0

Electrons used in an electron microscope are accelerated by a voltage of $$25$$ kV. If the voltage is increased to $$100$$ kV then the de-Broglie wavelength associated with the electrons would?

A
Increase by $$2$$ times

B
Decrease by $$2$$ times

C
Decrease by $$4$$ times

D
Increase by $$4$$ times

Solution

De Broglie wavelength in terms of accelerated potential difference is
$$\lambda =\dfrac{12.27}{\sqrt{V}}$$
Initial and final wavelength are $${{\lambda }_{1}}$$and $${{\lambda }_{2}}$$having potential as $${{V}_{1}}$$and $${{V}_{2}}$$
$$ \dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\sqrt{\dfrac{{{V}_{2}}}{{{V}_{1}}}} $$
$$ \dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\sqrt{\dfrac{100}{25}} $$
$$ \dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\sqrt{2} $$
$$ \dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=2 $$
$$ {{\lambda }_{2}}=\dfrac{{{\lambda }_{1}}}{2} $$
So, new wavelength is decrease by 2.
Question 9
1 / -0

The de Broglie wavelength $$\lambda $$ associated with a proton increases by $$25\%$$. If its momentum is decreased by $$p_0$$. The initial momentum was:

A
$$4 p_0$$

B
$$\dfrac{p_0}{4}$$

C
$$5p_0$$

D
$$\dfrac{p_0}{5}$$

Solution

Initial wavelength =$$\lambda$$
final =$$\lambda +\dfrac{\lambda}{4}=\dfrac{5 \lambda }{4}$$
Initial momentum = $$\dfrac{h}{\lambda}$$
final momentum = $$\dfrac{4\lambda}{5 \lambda}$$
$$\Delta P=\dfrac{h}{5 \lambda}=P_0$$
initial momentum =$$\dfrac{h}{\lambda}=5P_0$$
Question 10
1 / -0

The de-Broglie wavelength of an electron in the first Bohr orbit is

A
Equal to one fourth the circumference of the first orbit

B
Equal to twice the circumference of the first orbit

C
Equal to half the circumference of the first orbit

D
Equal to the circumference of the first orbit

Solution

$$mvr_n=\dfrac{nh}{2\pi}$$

$$\Rightarrow pr_n=\dfrac{nh}{2\pi}$$

$$\Rightarrow \dfrac{h}{\lambda}\times r_n=\dfrac{nh}{2\pi}$$

$$\Rightarrow \lambda =\dfrac{2\pi r_n}{n}$$, for first orbit $$n=1$$ so $$\lambda =2\pi r_1$$
= circumference of first orbit

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Dual Nature of Radiation and Matter Test - 45

Result

Dual Nature of Radiation and Matter Test - 45

Score

-

Rank

-

SHARING IS CARING

Question 1

h

h/e

eh

e

Solution

Question 2

$$n$$

$$\cfrac{1}{n}$$

$$n^2$$

$$\cfrac{1}{n^2}$$

Solution

Question 3

$$2.04\ \mathring {A}$$

$$1.04\ \mathring {A}$$

$$3.04\ \mathring {A}$$

$$4.04\ \mathring {A}$$

Solution

Question 4

For an electron as well as for a photon

For an electron but not for a photon

For an photon but not for an electron

Neither for an electron nor for a photon

Question 5

$$1.2 \mathring { A }$$

$$10 \mathring { A }$$

$$100 \mathring { A }$$

$$1 \mathring { A }$$

Solution

Question 6

1.55 eV

3.35 eV

4.85 eV

5.25 eV

Solution

Question 7

$$0.12$$ $$A^{\circ}$$

$$1.2$$ $$A^{\circ}$$

$$12.2$$ $$A^{\circ}$$

none of these.

Solution

Question 8

Increase by $$2$$ times

Decrease by $$2$$ times

Decrease by $$4$$ times

Increase by $$4$$ times

Solution

Question 9

$$4 p_0$$

$$\dfrac{p_0}{4}$$

$$5p_0$$

$$\dfrac{p_0}{5}$$

Solution

Question 10

Equal to one fourth the circumference of the first orbit

Equal to twice the circumference of the first orbit

Equal to half the circumference of the first orbit

Equal to the circumference of the first orbit

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.