NCERT

Yes, in most subjects, MPBSE follows NCERT textbooks or similar content. Subjects like Science, Math, and Social Science are mostly based on NCERT patterns. However, it's important to check the syllabus and match the chapters. Some subjects may include state-specific content, so students should refer to MP Board books too. A mix of NCERT and MPBSE resources gives complete coverage. Also, solve questions from both books for better results.

Xenon form compounds due to the high electronegativity of fluorine and oxygen, low ionization energy compared to lighter noble gases, and the availability of empty d-orbitals. This breaks the old assumption about noble gases being entirely inert.

With the combination of two different halogens, the interhalogen compounds are formed. When compared with pure diatomic halogen molecules, they are more reactive because the bond between different halogens are more polar and weaker. During reactions, it makes them more susceptible to bond cleavage.

Linear motion is one-dimensional motion. It refers to motion in a single direction or in a straight line. In linear motion, the object either moves forward or backward along one axis, i.e. x-axis. For example - a ball dropped from a height vertically downward or a car moving straight on a road. Motion in a Plane refers to an object moving in two dimensions, usually along x and y axes. For example, a football kicked at an angle.

The angle of projection is used to find the trajectory, horizontal range of a projectile, maximum height, and time of flight. For example, the maximum range on level ground is given by the 45-degree angle.

The vectors like velocity, displacement, and acceleration act along different directions in the two-dimensional motion. Resolving the vectors into the vertical and horizontal components allows the application of one-dimensional kinematic equations in each direction separately. It helps solve the problems more accurately and also simplifies the analysis.

Explanation- velocity of a freely falling body is v= $\sqrt[]{2gh}$

And $?=\frac{h}{mv}=\frac{h}{m\sqrt[]{2gh}}$

$?=h$ ^-1

Explanation-number of photon emitted per second n= $\frac{p}{\frac{hc}{?}}=\frac{p?}{hc}=\frac{20*5000*{10}^{-10}}{6.62*{10}^{-34}*3*{10}^8}$

$=5*{10}^{19}{s}^{-1}$

(ii) E=hc $/?$ = $\frac{6.62*{10}^{-34}*3*{10}^8}{5000*{10}^{-10}*1.6*{10}^{-19}}=2.48eV$  this enegy is greater than 2 so emission is possible

(iii) work function $?$ = $\frac{p}{4?d^2}*?r^2?t$ = ${?}_o$

$?t$ = $\frac{4?d^2}{p{r}^2}$ = $\frac{4*2*16*1.6*{10}^{-19}*2^2}{20*{(1.5*{10}^{-10})}^{-2}}=28.4s$

(iv) N= $\frac{n?r^2}{4?d^2}*?t$

 = $\frac{5*{10}^{19}*{(1.5*{10}^{-10})}^2*28.4}{4*({2)}^2}$ =2

(v) as the time of emission is 11.04s so photoelectric is not spontaneous.

Explanation- according to law of conservation of momentum

S0 m_Av+m_b0=m_Av₁+m_Bv₂

So m_A(v-v₁)= m_Bv₂

according to law of conservation of kinetic energy

1/2m_Av²=1/2m_Av₁²+1/2m_Bv₂²

So m_A(v²-v₁²)= m_Bv₂²

From above eqn we can say that v+v₁=v₂ or v=v₂-v₁

So v₁= $\frac{m_A{-m}_B}{m_A{+m}_B}$ v  and v₂= $\frac{2m_A}{m_A{+m}_B}$ v

_$?$ initial=h/m_Av

_$?$ final=h/m_Av₁= $\frac{h\left(m_A{+m}_B\right)}{m_{a\left(m_A{-m}_B\right)v}}$

$d?$ = $?$ _final- $?$ _initial= $\frac{h}{m_{A}v}\{\frac{m_A{+m}_B}{m_A{-m}_B}-1\}$

Explanation -Given threshold frequency of A is given by v_0A= 5 $*{10}^{14}$ hz

V_OB= 10 $*$ 10¹⁴hz

$?=h{v}_0$

$\frac{{?}_{OA}}{{?}_{OB}}=\frac{5*{10}^{14}}{10*{10}^{14}}1$

${?}_{OA}$ < ${?}_{OB}$

$\left(ii\right)$  for metal A slope=h/e= $\frac{2}{(10-5){10}^{14}}$

$h=\frac{2e}{5*{10}^{14}}=\frac{2*1.6*{10}^{-19}}{5*{10}^{14}}$ = 6.4 $*{10}^{-34}$ js

$\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{B}$ slope=h/e= $\frac{2.5}{(15-10){10}^{14}}$ = 8 $*{10}^{-34}$ js