Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 47

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Weekly Quiz Competition

Question 1
1 / -0

The radio of deBroglie wavelengths of a proton and an alpha particle of same energy is

A
$$1$$

B
$$2$$

C
$$4$$

D
$$0.25$$

Solution
Question 2
1 / -0

The de-Broglie wavelength of an electron is the same as the wavelength of a photon. The K.E of photon is $$'x'$$ times the K.E of the electron,then $$'x'$$ is (m-mass electron ,h- planck's constant,c-velocity of light)

A
$$\dfrac{hc}{2\lambda m}$$

B
$$\dfrac{2m\lambda c}{h}$$

C
$$\dfrac{2\lambda c}{hm}$$

D
$$\dfrac{2\lambda m}{ch}$$

Solution

$$\dfrac{hc}{\lambda}=x\left(\dfrac{1}{2}mv^2\right)$$
$$\dfrac{hc}{\lambda}=x\left(\dfrac{1}{2}m\left(\dfrac{h}{m\lambda}\right)^2\right)^2$$
$$=\dfrac{hc}{\lambda}=x\left(\dfrac{nh^2}{2m^2\lambda 2}\right)$$
$$=x=\dfrac{hc}{\lambda}\times \dfrac{2m \lambda ^2}{h^2}$$
$$=x=\dfrac{2m\lambda c}{h}$$
Question 3
1 / -0

The de Broglie wavelength corresponding to the root-mean-square velocity of hydrogen molecules at temperature $$20^{o}\ C$$ is

A
$$3.3A^{o}$$

B
$$1.04A^{o}$$

C
$$126A^{o}$$

D
$$34A^{o}$$

Solution
Question 4
1 / -0

A proton is accelerated through $$225\, V$$. Its de Brogile wavelength is:

A
$$1.91\, pm$$

B
$$0.2\, pm$$

C
$$3\, pm$$

D
$$0.4 \,nm$$

Solution

De-Brogile wavelength $$(\lambda)$$ of a changed particle when accelerate through a potential $$v$$ is given by:
$$\lambda=\dfrac{h}{\sqrt{2mqv}}$$ [$$h:$$ Plank's constant]
in case of proton, $$m$$(man of proton)=$$1.67\times 10^{-27}Kg$$
$$q$$(charge of proton)=$$1.6\times 10^{-19}C$$
Thus it gives:-$$[Given,V-225v]$$
$$\lambda=\dfrac{6.62\times 10^{-34}}{\sqrt{2\times 1.67\times 10^{-27}\times 1.6\times 10^{-19}\times 225}}$$
$$=\dfrac{6.62\times 10^{-34}}{\sqrt{1.2\times 10^{-43}}}$$
$$\Rightarrow \lambda \simeq 1.91\times 10^{-12}m=1.91pm.$$
Question 5
1 / -0

Electron has energy of 100 $${ e }{V }$$ what will be its wavelength

A
$$1.2{ \mathring { A } }$$

B
$$10 { \mathring { A } }$$

C
$$100 { \mathring { A } }$$

D
$$1 \mathring { A } $$

Solution
Question 6
1 / -0

A photon and an electron both have wavelength $$1A^{o}$$. The ratio of energy of photon to that of electron is

A
$$1$$

B
$$0.012$$

C
$$82.7$$

D
$$10^{-10}$$

Solution
Question 7
1 / -0

The de Broglie wavelength of an electron moving with a velocity $$1.5 \times{ 10 }^{ 8 }{ ms }^{ -1 }$$ is equal to that a photon. The ratio of the kinetic energy of the electron to the that of the photon is

A
2

B
4

C
$$\dfrac { 1 }{ 2 }$$

D
$$\dfrac { 1 }{ 4 }$$

Solution
Question 8
1 / -0

Mark the correct statement: In photo electric effect-

A
electrons are emitted from metal surface when light falls on it.

B
the kinetic energy of photo electrons is more for light of longer wavelength in composition to that due to shorter wavelength.

C
both of the above

D
none of the above

Solution

Einstein's photoelectric equation states that:
$$hv=\phi +K_{max}$$ (v$$=$$ frequency of incident light, $$\phi=$$ Work function of metal)
($$K_{max}=$$ kinetic energy of most energetic electron)
$$\Rightarrow \dfrac{hc}{\lambda}=\phi +K_{max}$$
As $$\phi =$$ constant so $$K_{max}$$ decreases when $$\lambda$$ increases
But electrons will not be ejected till light of suitable frequency which is greater or equal to the threshold frequency falls on it.
Question 9
1 / -0

A photon and an electron both have wavelength $$1\mathring { A } $$. The ratio of energy of photon to that of electron is

A
1

B
0.012

C
82.7

D
$${ 10 }^{ -10 }$$

Solution
Question 10
1 / -0

An $$\alpha $$ particle is moving in a circular path of radius in the presence of magnetic field B. The de-Broglie wavelength associated with the particle will be $$(q\rightarrow charge\quad on\quad \alpha \quad particle)$$

A
$$\cfrac { hr }{ qB } $$

B
$$\cfrac { hB }{ qr } $$

C
$$\cfrac { h }{ qBr } $$

D
$$\cfrac { h{ r }^{ 2 } }{ qB } $$

Solution

We know,

Radius $$ = r = \dfrac{mv}{qB}$$

and, $$\lambda = \dfrac{h}{mv}$$

$$\Rightarrow \ \lambda = \dfrac{h}{qBr}$$

$$\therefore$$ De-Broglie wavelength $$= \lambda = \dfrac{h}{qBr}$$

Hence, option "C" is correct.