Tag: physics

Questions Related to physics

The unit of Planck's constant is equivalent to that of

  1. energy

  2. angular momentum

  3. velocity

  4. force


Correct Option: B
Explanation:

The unit for Planck's constant is $ kg m^{2} s^{-1}$
The SI unit for angular momentum is $ kg m^{2} s^{-1}$
Hence they have the same SI units.
Option B is correct.

If the radius of first Bohrs orbit is $x$, then de-Broglie wavelength of electron in 3rd orbit is nearly

  1. $2\pi$ $x$

  2. $6\pi$$x$

  3. $9x$

  4. $x/3$


Correct Option: B
Explanation:

Radius of 3rd orbit radius $=9x=n^{2}x$ (where $n=3$ )
Let de broglie wavelength be $\lambda $.
For the interference of the waves to be constructive,
$n\lambda =2\pi r$ ($r$ is radius of orbit)
$\Rightarrow \lambda =\dfrac{2\pi \times 9x}{3} $ (where, $\ n=3 $, the quantum state)
$\Rightarrow \lambda =6\pi x$

The circumference of the second orbit of an atom or ion having single electron ,is $4 \times10^{-9}$ m.The de-Brogile wavelength of electron revolving in this orbit should be

  1. $2\times 10^{-9}m$

  2. $4\times 10^{-9}m$

  3. $8\times 10^{-9}m$

  4. $1\times 10^{-9}m$


Correct Option: A
Explanation:

The circumference of the orbit $=4\times 10^{9}m$
The orbit number $=2$

$n\lambda =2\pi r$

$\Rightarrow \lambda =\dfrac{4\times 10^{9}}{2}m$

$\Rightarrow \lambda =2\times 10^{-9}m$

If the electron in hydrogen orbit jumps from third orbit to second orbit, the wavelength of the emitted radiation is given by

  1. $\lambda = \dfrac {R}{6}$

  2. $\lambda = \dfrac {5}{R}$

  3. $\lambda = \dfrac {36}{5R}$

  4. $\lambda = \dfrac {5R}{36}$


Correct Option: C
Explanation:

We know that
$\dfrac {1}{\lambda} = R \left (\dfrac {1}{n _{1}^{2}} - \dfrac {1}{n _{2}^{2}} \right )$
$\dfrac {1}{\lambda} = R\left (\dfrac {1}{2^{2}} - \dfrac {1}{3^{2}} \right ) \Rightarrow R \left (\dfrac {1}{4} - \dfrac {1}{9}\right )$
$\dfrac {1}{\lambda} = \left (\dfrac {9 - 4}{36}\right ) R = \dfrac {5R}{36} \Rightarrow \lambda = \dfrac {36}{5R}$

According to de-Broglie explanation of Bohr's second postulate of quantization, the standing particle wave on a circular orbit for $n = 4$ is given by

  1. $2 \pi {r} _{n} = {4}/{\lambda}$

  2. $\dfrac{2 \pi}{\lambda} = 4{r} _{n}$

  3. $2 \pi {r} _{n} = 4 \lambda$

  4. $\dfrac{\lambda}{2 \pi} = 4 {r} _{n}$


Correct Option: C
Explanation:

According to debroglie explanation of Bohr's second postulate, assumption is made that integral number of wavelengths must fir in the circumference of circular orbit. The integral multiple comes out to be the same as quantization number.

$2 \pi r _n = n \lambda$
For $n=4$, 
       $2 \pi r _n = 4 \lambda$

Choose the wrong statement

  1. p-type semi conductor is positively charged

  2. n-type semi conductor is negatively charged

  3. both p-type and n-type are electrically neutral

  4. none of these


Correct Option: C
Explanation:

p and n type materials are NOT positively and negatively charged. An n-type material by itself has mainly negative charge carriers (electrons) which are able to move freely, but it is still neutral because the fixed donor atoms, having donated electrons, are positive

In N -- type semi  - conductor current is due to

  1. Electrons

  2. Holes

  3. Electrons and holes

  4. none of these


Correct Option: A
Explanation:

In N-type semi-conductor majority carriers are electrons. So, when the voltage is applied across the semi-conductor electrons started to drift from negative to positive terminal as result current starts to flow in opposite direction of electrons.

Hence in N- type semi-conductor current is due to electrons.

Fermi level of energy of an intrinsic semiconductor lies

  1. in the middle of the forbidden gap.

  2. below the middle of forbidden gap.

  3. above the middle of forbidden gap.

  4. outside the forbidden gap.


Correct Option: A
Explanation:

Fermi energy is determined as the energy point where the probability of occupancy by an electron is exactly $50\ %$ or $0.5$, i.e., $\dfrac{1}{2}$. For the intrinsic semiconductor, since electrons and holes are always created in pairs, $n = p = ni$. Hence, there are equal number of holes and electrons in valence band and conduction band respectively. Therefore, the Fermi energy level lies in the middle of the forbidden gap, i.e., energy band gap.

In a p-type semiconductor, the acceptor valence band is

  1. close to the valence band of the host crystal

  2. close to conduction band of the host crystal

  3. below the conduction band of the host crystal

  4. above the conduction band of the host crystal


Correct Option: A
Explanation:

The acceptor valence band is close to the valence band of host crystal