Dual Nature of Radiation and Matter Test

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Dual Nature of Radiation and Matter Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

The photo-electrons emitted from a surface of sodium metal are such that

A
They all are of the same frequency

B
They have the same kinetic energy

C
They have the same de-Broglie wavelength

D
They have their speeds from zero to a certain maximum

Solution

The photo-electrons emitted from a surface of sodium metal are such that
they have the same de-Broglie wavelength as
$$\lambda = \dfrac h{\sqrt{2mE}}$$ where,
$$E= h(\nu-\nu _0)$$
Question 2
1 / -0

The log-log graph between the energy $$E$$ of an electron and its de-Broglie wavelength $$\lambda$$ will be

A

B

C

D

Solution

As we know,
$$\lambda =\dfrac h{\sqrt{2mE}}$$

taking log on both sides, we get,
$$log\lambda=logh-\dfrac 12 log 2m -\dfrac 12logE$$
hence, graph will be a straight line with negative slope.
option C is correct.
Question 3
1 / -0

Which one is correct

A
$$E^2=p^2c^2$$

B
$$E^2=p^2c$$

C
$$E^2=pc^2$$

D
$$E^2=p^2/c^2$$

Solution

Momentum $$p=\dfrac {E}{c}\Rightarrow E^2=p^2c^2$$
Question 4
1 / -0

By photoelectric effect, Einstien proved

A
$$E=hv$$

B
$$K.E=\dfrac 12 mv^2$$

C
$$E=mc^2$$

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

Solution

By photoelectric effect, Einstien proved:
$$E=h\nu $$
Question 5
1 / -0

According to Bohr's theory, the electron in orbits have definite energy values, then according to uncertainty principle, the life time of an excited state will be

A
Zero

B
Finite

C
$$10\ sec$$

D
Infinite

Solution

According to Bohr's theory $$\Delta E=0,$$ since $$\Delta E.\Delta t \ge h$$
$$\Rightarrow \Delta t\rightarrow \infty$$
Question 6
1 / -0

The figure shows different graph between stopping potential $$(V_0)$$ and frequency $$(v)$$ for photosensitive surface of cesium, potassium, sodium and lithium. The plots are parallel. Correct ranking of the targets according to their work function greatest first will be

A
$$(i) > (ii) > (iii) >(iv) $$

B
$$(i) > (iii) > (ii) >(iv) $$

C
$$(iv) > (iii) > (ii) >(i) $$

D
$$(i) = (iii) > (ii) =(iv) $$

Solution

The graph between $$v_0$$ and $$v$$ cut the $$v-axis$$ at $$v$$.
For the given graphs $$(v_0)_{(iv)} > (v_0)_{(iii)} > (v_0)_{(ii)} > (v_0)_{(i)}$$
$$\therefore (W_0)_(iv) > (W_0)_(iii) > (W_0)_(ii) > (W_0)_(i)$$
Question 7
1 / -0

An electron is moving through a field. It is moving (i) opposite an electric field (ii) perpendicular to a magnetic field as shown. For each situation the de-Broglie wave length of electron

A
Increasing, Increasing

B
Increasing,decreasing

C
Decreasing, same

D
same,same

Solution

$$\lambda=\dfrac{h}{mv}$$. Since $$v$$ increasing in case (i), but it is not changing in case (ii). Hence, in the first case de-Broglie wavelength will change, but it second case, it remain the same
Question 8
1 / -0

According to Uncertainty principle if position of a any particle is measured by cent percent value., then the uncertainty in its momentum will be :

A
zero

B
$$ \infty $$

C
$$ \approx h $$

D
Nothing can be say

Solution

From Heisenberg Uncertainty principle
$$ \Delta p_x \Delta x > \dfrac{h}{2 \pi} $$

For accuracy in position change $$ \Delta x = 0 $$
$$ \therefore \, \Delta p_x = \dfrac{h}{2 \pi \times 0} = \infty $$
Question 9
1 / -0

Find out the de-Broglie wavelength related to an electron of kinetic energy $$10$$ eV:

A
$$10 \mathring A $$

B
$$ 1227 \mathring A $$

C
$$ 0.10 \mathring A $$

D
$$ 3.9 \mathring A $$

Solution

$$ \lambda = \dfrac{h}{\sqrt{2mE}} $$

$$ \lambda = \dfrac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 10 \times 1.6 \times 10^{-19}}} $$

$$ = 3.9 \mathring A $$
Question 10
1 / -0

An electron and a proton are bound in a linear box of dimension $$10A $$. Then the uncertainties in their momentum is:

A
$$ 1 : 1 $$

B
$$ 1 : 1836 $$

C
$$ 1836 : 1 $$

D
Insufficient information

Solution

From de-Broglie wavelength.
The uncertainty in their momentum
$$ \Delta x \Delta p_x = \dfrac{h}{2 \pi} $$

$$ \therefore $$ $$ \Delta p_x = \dfrac{h}{2 \pi \times \Delta x} $$

$$ \therefore $$ $$ \dfrac{\Delta px_e}{\Delta px_p} = \dfrac{h}{2 \pi \Delta x_e} \times \dfrac{\Delta x_p \times 2 \pi}{h} $$

$$ \therefore $$ $$ \dfrac{\Delta px_e}{\Delta px_p} = \dfrac{\Delta x_p}{\Delta x_e} = \dfrac{10 \times 10^{-10}}{10 \times 10^{-10}} $$

$$ \dfrac{\Delta px_e}{\Delta px_p} = \dfrac{1}{1} $$