NEET Test Series from KOTA - 10 Papers In MS WORD
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Structure of Atom
238704
If two particles $A$ and $B$ are moving with the same velocity. But wavelength of $A$ is found to be double than that of $B$. Which of the following statements are correct?
1 Both A and B have same mass
2 Mass of $A$ is half that of $B$
3 Mass of $B$ is half that of $A$
4 Mass of $B$ is one-fourth that of $A$
Explanation:
: From the equation of De-Broglie $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Where $\lambda=$ wavelength Or $\lambda=\frac{\mathrm{h}}{\mathrm{m} . \mathrm{v}} \quad \mathrm{h}=$ Plank's constant $\mathrm{P}=$ Momentum For particle $\mathrm{A}-$ $2 \lambda=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{A}} \mathrm{v}} \ldots \ldots \text { (i) }$ For particle B - $\lambda=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{B}} \mathrm{v}} \ldots \ldots$ From equation (i) and (ii) $2=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{A}} \mathrm{v}} \times \frac{\mathrm{m}_{\mathrm{B}} \mathrm{v}}{\mathrm{h}}$ Or $\quad 2=\frac{\mathrm{m}_{\mathrm{B}}}{\mathrm{m}_{\mathrm{A}}}$ $\mathrm{m}_{\mathrm{A}}=\frac{\mathrm{m}_{\mathrm{B}}}{2}$
AP EAPCET 19-08-2021 Shift-I
Structure of Atom
238708
The energy of an electromagnetic radiation is $3 \times 10^{-12}$ ergs. What is the wavelength in nanometers? $h=6.625 \times 10^{-27}$ erg sec. $\mathrm{c}=3 \times 10^{10} \mathrm{~cm} \mathrm{sec}^{-1}$
238709
Calculate the velocity of an electron having wavelength of $0.15 \mathrm{~nm}$. Mass of an electron is $9.109 \times 10^{-28}$ grams. $\left(h=6.626 \times 10^{-27}\right.$ ergsecond).
238710
Which of the following relations is correct, if the wavelength $(\lambda)$ is equal to the distance travelled by the electron in one second? $h$ is the Planck's constant and $m$ is the mass of electron
1 $\lambda=\mathrm{h} / \mathrm{p}$
2 $\lambda=\mathrm{h} / \mathrm{m}$
3 $\lambda=\sqrt{\mathrm{h} / \mathrm{p}}$
4 $\lambda=\sqrt{\mathrm{h} / \mathrm{m}}$
Explanation:
: We know that- $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ Here, $\quad \mathrm{m}=$ Mass of electron $\mathrm{v}=$ Velocity of electron $\lambda=$ Wavelength of electron If wavelength $(\lambda)$ is equal to distance travelled by the electron in one second i.e. $\lambda=\mathrm{v}$ Put in equation (i), we get- $\begin{aligned} \lambda & =\frac{\mathrm{h}}{\mathrm{m} \lambda} \\ \Rightarrow \quad \lambda^2 & =\frac{\mathrm{h}}{\mathrm{m}} \\ \lambda & =\sqrt{\frac{\mathrm{h}}{\mathrm{m}}} \end{aligned}$
238704
If two particles $A$ and $B$ are moving with the same velocity. But wavelength of $A$ is found to be double than that of $B$. Which of the following statements are correct?
1 Both A and B have same mass
2 Mass of $A$ is half that of $B$
3 Mass of $B$ is half that of $A$
4 Mass of $B$ is one-fourth that of $A$
Explanation:
: From the equation of De-Broglie $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Where $\lambda=$ wavelength Or $\lambda=\frac{\mathrm{h}}{\mathrm{m} . \mathrm{v}} \quad \mathrm{h}=$ Plank's constant $\mathrm{P}=$ Momentum For particle $\mathrm{A}-$ $2 \lambda=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{A}} \mathrm{v}} \ldots \ldots \text { (i) }$ For particle B - $\lambda=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{B}} \mathrm{v}} \ldots \ldots$ From equation (i) and (ii) $2=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{A}} \mathrm{v}} \times \frac{\mathrm{m}_{\mathrm{B}} \mathrm{v}}{\mathrm{h}}$ Or $\quad 2=\frac{\mathrm{m}_{\mathrm{B}}}{\mathrm{m}_{\mathrm{A}}}$ $\mathrm{m}_{\mathrm{A}}=\frac{\mathrm{m}_{\mathrm{B}}}{2}$
AP EAPCET 19-08-2021 Shift-I
Structure of Atom
238708
The energy of an electromagnetic radiation is $3 \times 10^{-12}$ ergs. What is the wavelength in nanometers? $h=6.625 \times 10^{-27}$ erg sec. $\mathrm{c}=3 \times 10^{10} \mathrm{~cm} \mathrm{sec}^{-1}$
238709
Calculate the velocity of an electron having wavelength of $0.15 \mathrm{~nm}$. Mass of an electron is $9.109 \times 10^{-28}$ grams. $\left(h=6.626 \times 10^{-27}\right.$ ergsecond).
238710
Which of the following relations is correct, if the wavelength $(\lambda)$ is equal to the distance travelled by the electron in one second? $h$ is the Planck's constant and $m$ is the mass of electron
1 $\lambda=\mathrm{h} / \mathrm{p}$
2 $\lambda=\mathrm{h} / \mathrm{m}$
3 $\lambda=\sqrt{\mathrm{h} / \mathrm{p}}$
4 $\lambda=\sqrt{\mathrm{h} / \mathrm{m}}$
Explanation:
: We know that- $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ Here, $\quad \mathrm{m}=$ Mass of electron $\mathrm{v}=$ Velocity of electron $\lambda=$ Wavelength of electron If wavelength $(\lambda)$ is equal to distance travelled by the electron in one second i.e. $\lambda=\mathrm{v}$ Put in equation (i), we get- $\begin{aligned} \lambda & =\frac{\mathrm{h}}{\mathrm{m} \lambda} \\ \Rightarrow \quad \lambda^2 & =\frac{\mathrm{h}}{\mathrm{m}} \\ \lambda & =\sqrt{\frac{\mathrm{h}}{\mathrm{m}}} \end{aligned}$
238704
If two particles $A$ and $B$ are moving with the same velocity. But wavelength of $A$ is found to be double than that of $B$. Which of the following statements are correct?
1 Both A and B have same mass
2 Mass of $A$ is half that of $B$
3 Mass of $B$ is half that of $A$
4 Mass of $B$ is one-fourth that of $A$
Explanation:
: From the equation of De-Broglie $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Where $\lambda=$ wavelength Or $\lambda=\frac{\mathrm{h}}{\mathrm{m} . \mathrm{v}} \quad \mathrm{h}=$ Plank's constant $\mathrm{P}=$ Momentum For particle $\mathrm{A}-$ $2 \lambda=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{A}} \mathrm{v}} \ldots \ldots \text { (i) }$ For particle B - $\lambda=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{B}} \mathrm{v}} \ldots \ldots$ From equation (i) and (ii) $2=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{A}} \mathrm{v}} \times \frac{\mathrm{m}_{\mathrm{B}} \mathrm{v}}{\mathrm{h}}$ Or $\quad 2=\frac{\mathrm{m}_{\mathrm{B}}}{\mathrm{m}_{\mathrm{A}}}$ $\mathrm{m}_{\mathrm{A}}=\frac{\mathrm{m}_{\mathrm{B}}}{2}$
AP EAPCET 19-08-2021 Shift-I
Structure of Atom
238708
The energy of an electromagnetic radiation is $3 \times 10^{-12}$ ergs. What is the wavelength in nanometers? $h=6.625 \times 10^{-27}$ erg sec. $\mathrm{c}=3 \times 10^{10} \mathrm{~cm} \mathrm{sec}^{-1}$
238709
Calculate the velocity of an electron having wavelength of $0.15 \mathrm{~nm}$. Mass of an electron is $9.109 \times 10^{-28}$ grams. $\left(h=6.626 \times 10^{-27}\right.$ ergsecond).
238710
Which of the following relations is correct, if the wavelength $(\lambda)$ is equal to the distance travelled by the electron in one second? $h$ is the Planck's constant and $m$ is the mass of electron
1 $\lambda=\mathrm{h} / \mathrm{p}$
2 $\lambda=\mathrm{h} / \mathrm{m}$
3 $\lambda=\sqrt{\mathrm{h} / \mathrm{p}}$
4 $\lambda=\sqrt{\mathrm{h} / \mathrm{m}}$
Explanation:
: We know that- $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ Here, $\quad \mathrm{m}=$ Mass of electron $\mathrm{v}=$ Velocity of electron $\lambda=$ Wavelength of electron If wavelength $(\lambda)$ is equal to distance travelled by the electron in one second i.e. $\lambda=\mathrm{v}$ Put in equation (i), we get- $\begin{aligned} \lambda & =\frac{\mathrm{h}}{\mathrm{m} \lambda} \\ \Rightarrow \quad \lambda^2 & =\frac{\mathrm{h}}{\mathrm{m}} \\ \lambda & =\sqrt{\frac{\mathrm{h}}{\mathrm{m}}} \end{aligned}$
238704
If two particles $A$ and $B$ are moving with the same velocity. But wavelength of $A$ is found to be double than that of $B$. Which of the following statements are correct?
1 Both A and B have same mass
2 Mass of $A$ is half that of $B$
3 Mass of $B$ is half that of $A$
4 Mass of $B$ is one-fourth that of $A$
Explanation:
: From the equation of De-Broglie $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Where $\lambda=$ wavelength Or $\lambda=\frac{\mathrm{h}}{\mathrm{m} . \mathrm{v}} \quad \mathrm{h}=$ Plank's constant $\mathrm{P}=$ Momentum For particle $\mathrm{A}-$ $2 \lambda=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{A}} \mathrm{v}} \ldots \ldots \text { (i) }$ For particle B - $\lambda=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{B}} \mathrm{v}} \ldots \ldots$ From equation (i) and (ii) $2=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{A}} \mathrm{v}} \times \frac{\mathrm{m}_{\mathrm{B}} \mathrm{v}}{\mathrm{h}}$ Or $\quad 2=\frac{\mathrm{m}_{\mathrm{B}}}{\mathrm{m}_{\mathrm{A}}}$ $\mathrm{m}_{\mathrm{A}}=\frac{\mathrm{m}_{\mathrm{B}}}{2}$
AP EAPCET 19-08-2021 Shift-I
Structure of Atom
238708
The energy of an electromagnetic radiation is $3 \times 10^{-12}$ ergs. What is the wavelength in nanometers? $h=6.625 \times 10^{-27}$ erg sec. $\mathrm{c}=3 \times 10^{10} \mathrm{~cm} \mathrm{sec}^{-1}$
238709
Calculate the velocity of an electron having wavelength of $0.15 \mathrm{~nm}$. Mass of an electron is $9.109 \times 10^{-28}$ grams. $\left(h=6.626 \times 10^{-27}\right.$ ergsecond).
238710
Which of the following relations is correct, if the wavelength $(\lambda)$ is equal to the distance travelled by the electron in one second? $h$ is the Planck's constant and $m$ is the mass of electron
1 $\lambda=\mathrm{h} / \mathrm{p}$
2 $\lambda=\mathrm{h} / \mathrm{m}$
3 $\lambda=\sqrt{\mathrm{h} / \mathrm{p}}$
4 $\lambda=\sqrt{\mathrm{h} / \mathrm{m}}$
Explanation:
: We know that- $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ Here, $\quad \mathrm{m}=$ Mass of electron $\mathrm{v}=$ Velocity of electron $\lambda=$ Wavelength of electron If wavelength $(\lambda)$ is equal to distance travelled by the electron in one second i.e. $\lambda=\mathrm{v}$ Put in equation (i), we get- $\begin{aligned} \lambda & =\frac{\mathrm{h}}{\mathrm{m} \lambda} \\ \Rightarrow \quad \lambda^2 & =\frac{\mathrm{h}}{\mathrm{m}} \\ \lambda & =\sqrt{\frac{\mathrm{h}}{\mathrm{m}}} \end{aligned}$
