Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 33

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Question 1
1 / -0

If a photon and an electron have same de-broglie wavelength, then

A
Both have same kinetic energy

B
Proton has more K.E. than electron

C
Electron has more K.E. than proton

D
Both have same velocity

Solution

De Broglie wavelength is given by:

$$\lambda = \dfrac{h}{p}$$

$$ p = \sqrt{2Em}$$ Where E is kinetic energy.
Given that both the electron and proton have the same De Broglie wavelength.
Then $$p_{e} = p_{p}$$
$$E_{electron}m_{electron} = E_{proton}m_{proton}$$

$$\dfrac{E_{electron}}{E_{proton}} = \dfrac{m_{proton}}{m_{electron}}$$

and we know that:

$$ \dfrac{m_{proton}}{m_{electron}} > 1 $$

Hence $$E_{electron} > E_{proton}$$
Question 2
1 / -0

Einstein's work on photoelectric effect provided support for the equation

A
$$E=hv$$

B
$$E=mc^2$$

C
$$E=\dfrac{-Rhc}{n^2}$$

D
$$K.E.=\dfrac{1}{2}mv^2$$

Solution

Einstein's photo electric effect & compton effect establish particle nature of light. These effects can be explained only, when we assume that the light has particle nature (To explain, Interference & diffraction the light must have wave nature. It means that light has both particle and have nature, so it is called dual nature of light).
$$ KE_{max} = E_{photon} - W_o$$
The above equation supports:
$$ E_{photon}= hv$$
It proves that light is in the form of dicrete packets of energy and not wave.
Otherwise the light with lower frequency than the threshold could give enough energy(slowly accumulate) to the electrons to come out of the metal.
Hence this theory supports particle nature of light, as suggested by Einstein.
Question 3
1 / -0

De Broglie wavelength for a beam of electron having energy 100 eV is :

A
$$1.23 \overset {o}{A}$$

B
$$2.46 \overset {o}A$$

C
$$1.48 \overset {o}A$$

D
$$2.45 \overset {o}A$$

Solution

E = $$\displaystyle\frac{1}{2}$$mv$$^2$$ = 100 eV = 100 $$\times$$ 1.6 $$\times$$ 10$$^{-19}$$ J

v$$^2$$ = $$\displaystyle\dfrac{2 E}{m}$$

v = $$\sqrt{2mE}$$

$$\lambda$$ = $$\displaystyle\dfrac{h}{mv}$$ = $$\displaystyle\frac{h}{\sqrt{2mE}}$$ metre

$$\lambda$$ = $$\displaystyle\dfrac{6.6 \times 10^{-34}}{ (2 \times 9.1 \times 10^{-31} \times 100 \times 1.6 \times 10^{-19} ) ^{1/2}}$$

= $$1.23 \overset {o}{A}$$

Option A is correct.
Question 4
1 / -0

What will be de Broglie wavelength of an electron moving with a velocity of $$\displaystyle 1.2\times 10^{5}ms^{-1} $$?

A
$$\displaystyle 6.068\times 10^{-9}m $$

B
$$\displaystyle 3.133\times 10^{-37}m $$

C
$$\displaystyle 6.626\times 10^{-9}m $$

D
$$\displaystyle 6.018\times 10^{-7}m $$

Solution

$$\displaystyle \lambda =\dfrac{h}{mv}$$

$$\displaystyle \lambda =\dfrac{6.62\times 10^{-34}}{9.1\times 10^{-31}\times 1.2\times 10^{5}}$$

$$\displaystyle \lambda =6.062\times 10^{-9}\: \: \: m$$

option (A) is correct.
Question 5
1 / -0

The figure shows a plot of photocurrent v/s anode potential for a photo sensitive surface for three different radiations. Which one of the following is a correct statement?

A
curves (b) and (c) represent incident radiations of same frequency having same intensity

B
curves (a) and (b) represent incident radiations of different frequencies and different intensities

C
curves (a) and (b) represent incident radiations of same frequency but of different intensities

D
curves (b) and (c) represent incident radiations of different frequencies and different intensities

Solution

The energy of the incident photon is $$h\nu$$.

Hence frequency determines the energy of incident photons. This energy is imparted on the photosensitive surface to cause photo ejection. At the points where curve cuts the x axis, the incident photon energy is equal to the work potential of material. Hence energy, hence frequency, is same for radiation of curves a & b.

Intensity is the measure of number of photons striking per unit time. Thus they determine the amount of photocurrent produced. Thus it is same for b & c.
Question 6
1 / -0

In H-atom, if 'x' is the radius of the first Bohr orbit, de Broglie wavelength of an electron in 3rd orbit is :

A
$$3\pi x$$

B
$$6\pi x$$

C
$$\displaystyle \frac{9 x}{2}$$

D
$$\displaystyle \frac{x}{2}$$

Solution

For a hydrogen atom, radius of nth orbit, $$r_n =\dfrac {n^2h^2}{4π^2me^2Z}$$

Therefore,$$ \dfrac{r_1}{r_3} = \dfrac {1^2}{3^2}$$

$$ r_3 = 9r_1$$=$$ 9x$$

Now, according to De Broglie; angular momentum of electron in $$3rd$$ orbit is:

$$mvr_3 = \dfrac{3h}{2π} $$ or $$ \dfrac{h}{mv} = \dfrac{2πr_3}3$$

And

$$λ = \dfrac{h}{mv}$$ , where $$\lambda$$ ,is the de Broglie wavelength.

Therefore,

$$\lambda = \dfrac{2πr_3}3$$

= $$\dfrac{2π . 9x}{3}$$ = $$ 6πx$$

Hence the answer is 6$$\pi x$$.
Question 7
1 / -0

de Broglie wavelength for a beam of electron having energy 100 eV in $$A^0$$ is :

A
$$1.23 $$

B
$$2.46$$

C
$$1.48$$

D
$$2.45$$

Solution

E = $$\displaystyle\frac{1}{2}$$mv$$^2$$ = 100 eV = 100 $$\times$$ 1.6 $$\times$$ 10$$^{-19}$$ J

v$$^2$$ = $$\displaystyle\frac{2 E}{m}$$

v = ($$\displaystyle\frac{2 E}{m})^{(1/2)}$$ = $$\sqrt{2mE}$$

$$\lambda$$ = $$\displaystyle\frac{h}{mv}$$ = $$\displaystyle\frac{h}{\sqrt{2mE}}$$ metre

$$\lambda$$ = $$\displaystyle\frac{6.6 \times 10^{-34}}{(2 \times 9.1 \times 10^{-31} \times 100 \times 1.6 \times 10^{-19}}^{1/2}$$

=1.23 $$\times$$ 10$$^{-10}$$ m = $$1.23 A^0$$
Question 8
1 / -0

If an electron and a photon propagate in the form of waves having the same wavelength, it implies that they have the same :

A
energy

B
momentum

C
velocity

D
angular momentum

Solution

In wave mechanical model momentum is given by $$\lambda=\dfrac{h}{p}$$
If $$\lambda$$ is same then momentum will be same.
Question 9
1 / -0

The electricity produced by creating photo electrons is called :

A
Photoelectricity

B
Static electricity

C
Thermal electricity

D
Piezoelectricity

Solution
Question 10
1 / -0

A photon of frequency n causes photoelectric emission from a surface with threshold frequency v$$_o$$. the de Broglie wavelength $$\lambda$$ of the photoelectron emitted is given as:

A
$$\Delta$$n = $$\displaystyle\frac{h}{2m \lambda}$$

B
$$\Delta$$n = $$\displaystyle\frac{h}{ \lambda}$$

C
[$$\displaystyle\frac{1}{ v_o}$$ - $$\displaystyle\frac{1}{ v}$$] = $$\displaystyle\frac{mc^2}{ h}$$

D
$$\lambda$$ = $$\displaystyle\sqrt{\frac{h}{ 2m \Delta n}}$$

Solution

E$$_T =$$ IE + KE

or $$E_T =$$ Threshold E (or) Work function + KE

or ($$hv = hv$$$$_0$$ $$+ 1/2 mu^2$$)

$$hv = hv$$$$_0$$ + $$\displaystyle\frac{1}{2}$$mu$$^2$$

$$\displaystyle\frac{1}{2}$$mu$$^2$$ $$= h(v - v$$$$_0$$) $$= h$$$$\Delta n$$ .... (i)

($$\lambda$$ $$=$$ $$\displaystyle\frac{h}{mu}$$, $$u =$$ $$\displaystyle\frac{h}{m\lambda}$$)

Substitute the value of u in equation (i)

$$\displaystyle\frac{1}{2}m\times$$ $$\displaystyle\frac{h^2}{m^2 \lambda^2}$$ $$= h$$$$\Delta n$$

$$\therefore \lambda =$$ $$\sqrt{\displaystyle\frac{h}{2m \Delta n}}$$

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Dual Nature of Radiation and Matter Test - 33

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Dual Nature of Radiation and Matter Test - 33

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Question 1

Both have same kinetic energy

Proton has more K.E. than electron

Electron has more K.E. than proton

Both have same velocity

Solution

Question 2

$$E=hv$$

$$E=mc^2$$

$$E=\dfrac{-Rhc}{n^2}$$

$$K.E.=\dfrac{1}{2}mv^2$$

Solution

Question 3

$$1.23 \overset {o}{A}$$

$$2.46 \overset {o}A$$

$$1.48 \overset {o}A$$

$$2.45 \overset {o}A$$

Solution

Question 4

$$\displaystyle 6.068\times 10^{-9}m $$

$$\displaystyle 3.133\times 10^{-37}m $$

$$\displaystyle 6.626\times 10^{-9}m $$

$$\displaystyle 6.018\times 10^{-7}m $$

Solution

Question 5

curves (b) and (c) represent incident radiations of same frequency having same intensity

curves (a) and (b) represent incident radiations of different frequencies and different intensities

curves (a) and (b) represent incident radiations of same frequency but of different intensities

curves (b) and (c) represent incident radiations of different frequencies and different intensities

Solution

Question 6

$$3\pi x$$

$$6\pi x$$

$$\displaystyle \frac{9 x}{2}$$

$$\displaystyle \frac{x}{2}$$

Solution

Question 7

$$1.23 $$

$$2.46$$

$$1.48$$

$$2.45$$

Solution

Question 8

energy

momentum

velocity

angular momentum

Solution

Question 9

Photoelectricity

Static electricity

Thermal electricity

Piezoelectricity

Solution

Question 10

$$\Delta$$n = $$\displaystyle\frac{h}{2m \lambda}$$

$$\Delta$$n = $$\displaystyle\frac{h}{ \lambda}$$

[$$\displaystyle\frac{1}{ v_o}$$ - $$\displaystyle\frac{1}{ v}$$] = $$\displaystyle\frac{mc^2}{ h}$$

$$\lambda$$ = $$\displaystyle\sqrt{\frac{h}{ 2m \Delta n}}$$

Solution

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