Physics

13.6 In $\alpha$ - decay, there is a loss of 2 protons and 4 neutrons. In every $\beta +$ decay, there is a loss of 1 proton and a neutrino is emitted from the nucleus. In every $\beta -$ decay, there is a gain of 1 proton and an antineutrino is emitted from the nucleus.

${}_{88}{}^{226}Ra$ $\to$ ${}_{86}{}^{222}Rn$ + ${}_2{}^{4}He$

${}_{94}{}^{242}Pu$ $\to$ ${}_{92}{}^{238}U$ + ${}_2{}^{4}He$

${}_{15}{}^{32}P$ $\to$ ${}_{16}{}^{32}S$ + $e^{-}$ + $\stackrel{?}{\nu }$

${}_{83}{}^{210}Bi$ $\to$ ${}_{84}{}^{210}Po$ + $e^{-}$ + $\stackrel{?}{\nu }$

${}_6{}^{11}C$ $\to$ ${}_5{}^{11}B$ + $e^{+}$ + $\nu$

${}_{43}{}^{97}Tc$ $\to$ ${}_{42}{}^{97}Mo$ + $e^{+}$ + $\nu$

${}_{54}{}^{120}Xe\to$ ${}_{53}{}^{120}I$ + $\nu$

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.

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