# 11 - Modern Physics Questions Answers

A point source of light is placed at the centre of curvature

of a hemispherical surface. The radius of curvature is r

and the inner surface is completely reflecting. Find the

force on the hemisphere due to the light falling on it if

the source emits a power W.

**Asked By: ANKIT SHARMA**

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There are N numbers of electrons present in a one-dimensional box of length L. Assuming no coulombic interactions are present, find the ground-state energy of the N-particle system. N is an even number and mass of each electron is m. All electrons have a definite momentum, p, and their collisions with the box are perfectly elastic.

**Asked By: SAJAG KUMAR**

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**Joshi sir comment**

Sir

Does time exist or is it just an illusion .

Please tell.

**Asked By: SHIVAM SHARMA**

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**Joshi sir comment**

I am not a scientist yet many scientists including Einstein explain time as an illusion, so it can be said.

*Consider an electron trapped in an infinite well whose width is 98.5 pm. If it is in a state with n = 15, what are (i) its energy? (ii) the uncertainty in its momentum? (iii) the uncertainty in its position?*

**Asked By: ANKIT SONI**

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**Joshi sir comment**

$Letwidth=b\phantom{\rule{0ex}{0ex}}Fornthstaten\frac{\lambda}{2}=b\Rightarrow \lambda =\frac{2b}{n}\phantom{\rule{0ex}{0ex}}NowE=\frac{{p}^{2}}{2m}=\frac{{\left({\displaystyle \raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$\lambda $}\right.}\right)}^{2}}{2m}\phantom{\rule{0ex}{0ex}}Solveafterputtingthevalueof\lambda \phantom{\rule{0ex}{0ex}}Uncer\mathrm{ta}intyinposition=b\phantom{\rule{0ex}{0ex}}Nowuse\u2206p.\u2206x=nh/2\mathrm{\pi}\mathrm{for}\mathrm{finding}\mathrm{uncertainty}\mathrm{in}\mathrm{momentum}$

a toy car of mass 500g travels with a uniform velocity of 25m/s for 5 seconds, the brakes are then applied and the car is uniformly retarded and comes to rest in further 10s calculate the retardation

**Asked By: SARABJEET**

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**Joshi sir comment**

use v = u + at

u = 25, v = 0 and t = 10

when we throw a shotput ball it covers a very less distance and when we throw a rubber ball it covers a larger distance.

the reason behind this well known to everyone that the shotput ball has more mass as compared to the rubber ball.

**But when throw a paper ball it covers the least diatance why?**

**Asked By: SANTANU VERMA**

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**Joshi sir comment**

for this motion 0-R = ma

so retardation = R/m

for less mass retardation will be more

in the case of shotput

mu + impulse given = mv

so v = u + {impulsegiven/m}

so more mass will gain less velocity in comparison to less mass on applying same impulse.

the half life of radium is 1620 year & its atomic weight is 226gm/mole. the number of atoms that will decay from its 1g sample per second is

1) 3.16 x 10^{10}

2)3.6 x 10^{12}

3) 3.1 x 10^{15}

4) 31.1 x 10^{15}

**Asked By: HIMANSHU**

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**Joshi sir comment**

for radioactivity we know that it is a first order process

so dN/dt = -λN

so dN/1 = -[0.6932/(1620*3.1*10^{7})]*(1/226)*6.023*10^{23}

solve now

A RADIOACTIVE SUBSTANCE EMITS n BETA PARTICLES IN THE FIRST 2 SECONDS & 0.5 n BETA PARTICLES IN THE NEXT 2 SECONDS . THE MEAN LIFE OF THE SAMPLE IS

ANS 2/(ln2) s

**Asked By: SARIKA SHARMA**

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**Joshi sir comment**

A RADIOACTIVE SUBSTANCE EMITS n BETA PARTICLES IN THE FIRST 2 SECONDS & 0.5 n BETA PARTICLES IN THE NEXT 2 SECONDS. Definitely by definition its half life will be 2 sec.

so mean life = half life / ln2 This is the relation between mean life and half life.

A NUCLEUS OF MASS NO. 220, INITIALLY AT REST , EMITS AN .α PARTICLE . IF THE Q VALUE OF THE Rxn IS 5.5MeV THE ENERGY OF THE EMITTED α PARTICLE WILL BE ??

ANS 5.4 MeV

**Asked By: SARIKA SHARMA**

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**Joshi sir comment**

5.5*216/220 = 216/40 = 5.4 MeV here 216 is the mass of nucleus after ejecting α particle.

The activity of a freshly prepared radioactive sample is 10^10 disintegrations per second, whose mean life is 10^9 sec.

The mass of an atom of this radioisotope is 10 ^ -25 kg. The mass (in mg) of the radioactive sample is

**Asked By: NITIN RAO**

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**Joshi sir comment**

dN/dt = -λN

10^{10} = -[1/10^{9}] N λ = 1/mean life

so N = 10^{19 }disintegration

so mass of radioisotope = 10^{19} *10^{-25 }= 10^{-6} Kg = 1 mg