Physics

Explanation- given d= 0.1nm,  $?$ =30⁰ and n=1

According to bragg's law

2dsin $?$ = n $?$

2 $*0.1*sin30=?$

$?=0.1nm={10}^{-10}$

$?=\frac{h}{mv}=\frac{h}{p}$

P=h/ $?$  = $\frac{6.62*{10}^{-34}}{{10}^{-10}}$ = 6.62 $*{10}^{-24}$ kgm/s

k.E= 1/2mv²= $\frac{1}{2}\frac{m^2{v}^2}{m}$ = $\frac{p^2}{2m}$ = 0.21eV

The de Broglie wavelength  depends on:
Planck's constant h (a universal constant),
velocity v of the particle
mass m of the particle.

Photons do not have enough energy ( )to overcome the work function (? ) of the material below the threshold frequency.The energy per photon remains too low to liberate electrons even if the light intensity is increased, so photoelectric emission cannot occur.

According to de Broglie's hypothesis, all matter exhibits wave-like behavior. It laid the foundation for quantum mechanics by introducing the concept of matter waves. The dual nature of particles (both wave and particle) is essential in designing technologies like electron microscopes and helps explain phenomena like electron diffraction.

The NCERT Exemplar given on Shiksha's website provides a variety of numerical, conceptual and application-based questions around energy transitions, atomic models, and spectral series. It is a powerful study material for CBSE Board examination preparation and competitive exams like NEET and JEE.

According to Bohr's postulates, in specific stable orbits, electrons revolve without emitting energy. Energy is only absorbed or emitted when electrons jump between these orbits, preventing them from spiraling into the nucleus.

The concept of quantized energy levels for electrons in atoms is introduced by Bohr's model. Although, Bohr's model could not explain the spectra of multi-electron atoms. It helped lay the foundation for quantum mechanics and successfully explained the spectral lines of hydrogen.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Explanation- |A+B|=|A-B|

$\sqrt[]{\left|{A|}^2+\left|{B|}^2+2\right|A\right|\left|B\right|cos\theta }=\sqrt[]{\left|{A|}^2+\left|{B|}^2-2\right|A\right|\left|B\right|cos\theta }$

$\left|{A|}^2+\left|{B|}^2+2\right|A\right|\left|B\right|cos\theta$  = $\left|{A|}^2+\left|{B|}^2-2\right|A\right|\left|B\right|cos\theta$

4|A|B|cos $\theta$ =0

|A|²+|B|²cos $\theta$ =0

A=0 or B=0 so $\theta =90$ . so A perpendicular B

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, b, c

Explanation- (i) speed will constant throughout

(ii) velocity will be tangential in the direction of motion

(iii) centripetal acceleration will be a= v²/r, will always be towards centre of the circular path.

(iv) angular momentum is constant in magnitude and direction out of the plane perpendicularly as well.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, c

Explanation – as we know average acceleration is a_av= $\frac{?v}{?t}=\frac{v_2-v_1}{t_2-t_1}$

But when acceleration is not uniform V_av is not equal to v₁+v₂/2

So we can write $?v=\frac{?r}{?t}$

$?r=r_2-r_1$ = v₂-v₁ (t₂-t₁)