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2T

T

3/2T

3T

Radioactive decay, $N(t) = N_{0} e^{- \frac{0.693t}{t_{1/2}}}$

Where N(t) is the amount of sample which remains after time t.

Given that 75% of the sample has decayed, hence the remaining will be 25% of initial, $N(t) =0.25 N_{0}$

Also $t_{1/2} =T$

$0.25 N_{0} =N_{0}e^{- \frac{0.693t}{T}}$

$\Rightarrow 0.25 =e^{- \frac{0.693t}{T}}$

Take loge both sides and solve

$t = 2T$

Option (1) is correct

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Posted by

Abhishek Sahu

When a radioactive isotope ₈₈Ra²²⁸ decays in series by the emission of three α-particles and a β particle the isotope finally formed is :

Option 1)
$_{84}X^{220}$

Option 2)
$_{86}X^{222}$

Option 3)
$_{83}X^{216}$

Option 4)
$_{83}X^{215}$

As learnt in

? -decay -

$^{A}_{Z}X\rightarrow ^{A-4}_{Z-2}Y+^{4}_{2}He+Q$

- wherein

$Q\: value = \left ( M_{X} -M_{Y}-M_{He}\right )c^{2}$

Option 1)

$_{84}X^{220}$

This option is incorrect

Option 2)

$_{86}X^{222}$

This option is incorrect

Option 3)

$_{83}X^{216}$

This option is correct

Option 4)

$_{83}X^{215}$

This option is incorrect

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Posted by

prateek

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The X-ray beam emerging from an X-ray tube

Option 1)
is monochromatic

Option 2)
contains all wavelengths smaller than a certain maximum wavelength

Option 3)
contains all wave lengths larger than a certain minimum wavelength

Option 4)
contains all wave lengths lying between a minimum and a maximum wavelength.

As learnt in

X-Ray -

Energy produced when higher atomic number atoms make transition

- wherein

It is of two types: continuous & characteristics X- Ray.

X-ray has both continuous and characteristics wavelength. It contains wavelength from minimum to a large value.

Option 1)

is monochromatic

This option is incorrect

Option 2)

contains all wavelengths smaller than a certain maximum wavelength

This option is incorrect

Option 3)

contains all wave lengths larger than a certain minimum wavelength

This option is correct

Option 4)

contains all wave lengths lying between a minimum and a maximum wavelength.

This option is incorrect

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Posted by

Aadil

The wave length λ of Kα-ray line of an anticathode element of atomic number Z is nearly proportional to:

Option 1)
$Z^{2}$

Option 2)
$\left ( Z-1 \right )^{2}$

Option 3)
$\frac{1}{\left ( Z-1 \right )}$

Option 4)
$\frac{1}{\left ( Z-1 \right )^{2}}$

Moseley's law -

$\sqrt{\nu }=a(z-b)$

- wherein

$a=\sqrt{\frac{3RC}{4} }$

b=1 for

$K_{\alpha} \, \, lines$

$\sqrt{\nu }= a\left ( z-1 \right )\left [ b=1 \right ]or \sqrt{\frac{c}{\lambda }}=a\left ( z-1 \right ) or \frac{c}{\lambda }= a^{2}\left ( z-1 \right )^{2}$

$\therefore \lambda \alpha \frac{1}{(Z-1)^{2}}$

Option 1)

$Z^{2}$

Incorrect option

Option 2)

$\left ( Z-1 \right )^{2}$

Incorrect option

Option 3)

$\frac{1}{\left ( Z-1 \right )}$

Incorrect option

Option 4)

$\frac{1}{\left ( Z-1 \right )^{2}}$

correct option

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Posted by

Plabita

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

The total energy of the electron in the hydrogen atom in the ground state is -13.6 eV. Which of the following is its kinetic energy in the first excited state?

Option 1)
13.6 eV

Option 2)
6.8 eV

Option 3)
3.4 eV

Option 4)
1.825 eV

Relation between K.E.,P.E. & total energy -

$K.E.= \left ( Total\: energy \right )\\=\frac{1}{2}\left ( Potential \: energy \right )$

- wherein

Potential Energy and total energy are negative .While kinetic energy is positive

Total energy in first excited state = $\dot-13.6 (\frac{1}{4}) =-3.4 ev$

$K.E=\left | total \ energy \right |=3.4 ev$

Option 1)

13.6 eV

Incorrect option

Option 2)

6.8 eV

Incorrect option

Option 3)

3.4 eV

correct option

Option 4)

1.825 eV

Incorrect option

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Posted by

Vakul

The stopping potential for the photo electrons emitted from a metal surface of work function 1.7 eV is 10.4 V. Identify the energy levels corresponding to the transitions in hydrogen atom which will result in emission of wavelength equal to that of incident radiation for the above photoelectric effect

Option 1)
n = 3 to 1

Option 2)
n = 3 to 2

Option 3)
n = 2 to 1

Option 4)
n = 4 to 1

Stopping Potential /Cut-off Potential -

It is defined as the potential necessary to stop any electron from reaching the other side.

Conservation of energy -

$h\nu = \phi _{0}+\frac{1}{2}mv^{2}_{max}$

$h\nu = h\nu _{0}+\frac{1}{2}mv^{2}_{max}$

$h\left ( \nu-\nu _{0} \right )=\frac{1}{2}mv^{2}_{max}$

$where, h - Plank's constant\ \nu - Frequency\ \nu_{0} - threshold\ frequency\ \phi_{0}- work function$

- wherein

$eV_{s}=K.E. - \phi .-(i)$

$\because V_{s}=$ stopping potential

$\Rightarrow 10.4 =K.E-1.7$

$\therefore K.E = 12.1 ev$

This energy can be provided by the transition of an electron from n=3 to n=1

$\\ \Delta E = 13.6 (\frac{1}{12}-\frac{1}{9})=12.1 ev \\ 3\rightarrow 1$

Option 1)

n = 3 to 1

Correct option

Option 2)

n = 3 to 2

Incorrect option

Option 3)

n = 2 to 1

Incorrect option

Option 4)

n = 4 to 1

Incorrect option

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Posted by

Vakul

NEET 2024 Most scoring concepts

Just Study 32% of the NEET syllabus and Score up to 100% marks

The relation between half-life T of a radioactive sample and its mean life τ is:

Option 1)
T = 0.693 τ

Option 2)
τ = 0.693 T

Option 3)
τ = T

Option 4)
τ = 2.718 T

Average life -

$T_{av}= \frac{1}{\lambda }$

- wherein

It is defined as time in which number of nuclei reduces to $\left ( \frac{1}{e} \right )$ part of initial number of nuclei

Half Life Time -

$t_{1/2}=\frac{ln 2}{\lambda}=\frac{0.693}{\lambda}$

- wherein

Half life is time in which number of nuclei reduced to half of initial number of nuclei.

Half life T = $\frac{D.693}{\lambda }$

Average life (mean life) I = $\frac{1}{\lambda }$

Option 1)

T = 0.693 τ

Incorrect option

Option 2)

τ = 0.693 T

Incorrect option

Option 3)

τ = T

correct option

Option 4)

τ = 2.718 T

Incorrect option

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Posted by

Plabita

The ratio of magnetic dipole moment of an electron of charge e and mass m in the Bohr orbit in hydrogen to the angular momentum of the electron in the orbit is:

Option 1) e/m

Option 2) e/2m

Option 3) m/e

Option 4) 2m/e

Bohr quantisation principle -

$mvr=\frac{nh}{2\pi } \\2\pi r= n\lambda$

- wherein

Angular momentum of an electron in stationary orbit is quantised.

magnetic dipole moment, $(\mu )$ $=IA$

$=\left ( \frac{e}{T} \right )\cdot \pi r^{2}= \frac{e}{\left ( \frac{2\pi r}{v} \right )}\cdot \pi r^{2} = \frac{evr}{2}$

angular momentum of electron $L=mvr$

$\therefore \frac{\mu }{L} = \frac{evr}{2\cdot mvr} = \frac{e}{2m}$

Option 1)

$\frac{e}{m}$

This solution is incorrect.

Option 2)

$\frac{e}{2m}$

This solution is correct.

Option 3)

$\frac{m}{e}$

This solution is incorrect.

Option 4)

$\frac{2m}{e}$

This solution is incorrect.

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Posted by

Aadil

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The radius of the first Bohr orbit is a_o. The nth orbit has a radius:

Option 1)
na_o

Option 2)
a_o/n

Option 3)
n²a_o

Option 4)
a_o/n²

Radius of nth orbital -

$r_{n}= \frac{\epsilon _{0}n^{2}h^{2}}{\pi mZe^{2}}$

- wherein

$r_{n}\alpha \: \frac{n^{2}}{Z}$

$\frac{\epsilon_{0}h^{2}}{\pi me^{2}}= 0.529A^{\circ}$

$r=\frac{n^{2}}{z} a_{0}$

For $z=1, r=n^{2}a_{0}$

Option 1)

na_o

This solution is incorrect.

Option 2)

a_o/n

This solution is incorrect.

Option 3)

n²a_o

This solution is correct.

Option 4)

a_o/n²

This solution is incorrect.

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Posted by

divya.saini

The potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation:

Option 1)
the maximum wavelength increases

Option 2)
the minimum wave length increases

Option 3)
the minimum wavelength remains unchanged

Option 4)
the minimum wave length decreases

Continuous x-ray -

$\lambda ^{min}=\frac{hc}{ev}$

- wherein

Also called cutoff wavelength . All other wavelength are greater than this.

For X ray.

$\lambda ^{min}=\frac{hc}{ev} \Rightarrow \lambda ^{min} \alpha ^{-} \frac{1}{v}$

As we increase the applied potential difference, $\lambda ^{min}$ will decrease.

Option 1)

the maximum wavelength increases

This solution is incorrect.

Option 2)

the minimum wave length increases

This solution is incorrect.

Option 3)

the minimum wavelength remains unchanged

This solution is incorrect.

Option 4)

the minimum wave length decreases

This solution is correct.

View Full Answer(1)

Posted by

prateek

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Get Answers to all your Questions

All Questions

2T T 3/2T 3T

Posted by

Abhishek Sahu

When a radioactive isotope 88Ra228 decays in series by the emission of three α-particles and a β particle the isotope finally formed is : Option 1) Option 2) Option 3) Option 4)

Posted by

prateek

Crack CUET with india's "Best Teachers"

Posted by

Aadil

The wave length λ of Kα-ray line of an anticathode element of atomic number Z is nearly proportional to: Option 1) Option 2) Option 3) Option 4)

Posted by

Plabita

JEE Main high-scoring chapters and topics

The total energy of the electron in the hydrogen atom in the ground state is -13.6 eV. Which of the following is its kinetic energy in the first excited state? Option 1) 13.6 eV Option 2) 6.8 eV Option 3) 3.4 eV Option 4) 1.825 eV

Posted by

2T

T

3/2T

3T

When a radioactive isotope ₈₈Ra²²⁸ decays in series by the emission of three α-particles and a β particle the isotope finally formed is :

Option 1)
$_{84}X^{220}$

Option 2)
$_{86}X^{222}$

Option 3)
$_{83}X^{216}$

Option 4)
$_{83}X^{215}$

The wave length λ of Kα-ray line of an anticathode element of atomic number Z is nearly proportional to:

Option 1)
$Z^{2}$

Option 2)
$\left ( Z-1 \right )^{2}$

Option 3)
$\frac{1}{\left ( Z-1 \right )}$

Option 4)
$\frac{1}{\left ( Z-1 \right )^{2}}$

The total energy of the electron in the hydrogen atom in the ground state is -13.6 eV. Which of the following is its kinetic energy in the first excited state?

Option 1)
13.6 eV

Option 2)
6.8 eV

Option 3)
3.4 eV

Option 4)
1.825 eV

The relation between half-life T of a radioactive sample and its mean life τ is:

Option 1)
T = 0.693 τ

Option 2)
τ = 0.693 T

Option 3)
τ = T

Option 4)
τ = 2.718 T

The ratio of magnetic dipole moment of an electron of charge e and mass m in the Bohr orbit in hydrogen to the angular momentum of the electron in the orbit is:

Option 1) e/m

Option 2) e/2m

Option 3) m/e

Option 4) 2m/e

The radius of the first Bohr orbit is a_o. The nth orbit has a radius:

Option 1)
na_o

Option 2)
a_o/n

Option 3)
n²a_o

Option 4)
a_o/n²

The potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation:

Option 1)
the maximum wavelength increases

Option 2)
the minimum wave length increases

Option 3)
the minimum wavelength remains unchanged

Option 4)
the minimum wave length decreases