Question:
A particle of mass
\[
2\times10^{-27}\,\text{kg}
\]
has de-Broglie wavelength of
\[
3.3\times10^{-10}\,\text{m}.
\]
The kinetic energy of this particle is:
\[
\text{(Planck’s constant } h=6.6\times10^{-34}\,\text{J-s)}
\]
A particle of mass
\[
2\times10^{-27}\,\text{kg}
\]
has de-Broglie wavelength of
\[
3.3\times10^{-10}\,\text{m}.
\]
The kinetic energy of this particle is:
\[ \text{(Planck’s constant } h=6.6\times10^{-34}\,\text{J-s)} \]
\[ \text{(Planck’s constant } h=6.6\times10^{-34}\,\text{J-s)} \]
Show Hint
For de-Broglie wavelength problems, directly use:
\[
K=\frac{h^2}{2m\lambda^2}
\]
to save calculation time.
Updated On: Jun 24, 2026
- \(5\times10^{-20}\,\text{J}\)
- \(8\times10^{-20}\,\text{J}\)
- \(1\times10^{-21}\,\text{J}\)
- \(6\times10^{-22}\,\text{J}\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Use de-Broglie wavelength relation.
According to de-Broglie hypothesis, \[ \lambda=\frac{h}{p} \] where \[ p=mv \] Hence, \[ mv=\frac{h}{\lambda} \]
Step 2: Write the expression for kinetic energy.
Kinetic energy is \[ K=\frac{1}{2}mv^2 \] Using \[ mv=\frac{h}{\lambda}, \] we get \[ K=\frac{1}{2m}\left(\frac{h}{\lambda}\right)^2 \] Thus, \[ K=\frac{h^2}{2m\lambda^2} \]
Step 3: Substitute the given values.
\[ h=6.6\times10^{-34}\,\text{J-s} \] \[ m=2\times10^{-27}\,\text{kg} \] \[ \lambda=3.3\times10^{-10}\,\text{m} \] Therefore, \[ K= \frac{(6.6\times10^{-34})^2} {2(2\times10^{-27})(3.3\times10^{-10})^2} \]
Step 4: Simplify the expression.
\[ (6.6)^2=43.56 \] \[ (3.3)^2=10.89 \] \[ K= \frac{43.56\times10^{-68}} {4\times10.89\times10^{-47}} \] \[ K= \frac{43.56}{43.56}\times10^{-21} \] \[ K=10^{-21}\,\text{J} \]
Step 5: Final conclusion.
Hence, the kinetic energy of the particle is \[ \boxed{1\times10^{-21}\,\text{J}} \]
According to de-Broglie hypothesis, \[ \lambda=\frac{h}{p} \] where \[ p=mv \] Hence, \[ mv=\frac{h}{\lambda} \]
Step 2: Write the expression for kinetic energy.
Kinetic energy is \[ K=\frac{1}{2}mv^2 \] Using \[ mv=\frac{h}{\lambda}, \] we get \[ K=\frac{1}{2m}\left(\frac{h}{\lambda}\right)^2 \] Thus, \[ K=\frac{h^2}{2m\lambda^2} \]
Step 3: Substitute the given values.
\[ h=6.6\times10^{-34}\,\text{J-s} \] \[ m=2\times10^{-27}\,\text{kg} \] \[ \lambda=3.3\times10^{-10}\,\text{m} \] Therefore, \[ K= \frac{(6.6\times10^{-34})^2} {2(2\times10^{-27})(3.3\times10^{-10})^2} \]
Step 4: Simplify the expression.
\[ (6.6)^2=43.56 \] \[ (3.3)^2=10.89 \] \[ K= \frac{43.56\times10^{-68}} {4\times10.89\times10^{-47}} \] \[ K= \frac{43.56}{43.56}\times10^{-21} \] \[ K=10^{-21}\,\text{J} \]
Step 5: Final conclusion.
Hence, the kinetic energy of the particle is \[ \boxed{1\times10^{-21}\,\text{J}} \]
Was this answer helpful?
0
0
Top AP EAPCET Physics Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is $ 0{}^\circ $ ) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is: (g = acceleration due to gravity)
- A rigid metallic sphere is spinning around its own axis in the absence of external torque. If the temperature is raised, its volume increases by $9 \%$. The change in its angular speed is
View More Questions
Top AP EAPCET de broglie hypothesis Questions
- An $\alpha$-particle moves in a circular path of radius $1\, cm$ in a uniform magnetic field of $0.125\, T$. The de Broglie wavelength associated with the $\alpha$-particle is
- An electron, a proton and an alpha particle are moving with same kinetic energy. If $ \lambda_1, \lambda_2, \lambda_3 $ are the de Broglie wavelengths of the electron, proton and alpha particle respectively, then
- The de Broglie wavelength of a charged particle accelerated through a potential difference $ V $ is $ \lambda $. If the potential difference is increased by 21%, the de Broglie wavelength of the charged particle is:
- The de Broglie wavelength of an electron in the first orbit of hydrogen atom is:
- The velocity of a particle A is 3 times the velocity of proton. If the ratio of the de Broglie wavelengths of the particle A and the proton is 3:2, the mass of the particle A is:
View More Questions
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions