The energy equivalent of 1g of substance is:
- \( 11.2 \times 10^{24} \, \text{MeV} \)
- \( 5.6 \times 10^{12} \, \text{MeV} \)
- \( 5.6 \, \text{eV} \)
- \( 5.6 \times 10^{26} \, \text{MeV} \)
The Correct Option is D
Approach Solution - 1
To find the energy equivalent of 1 gram of a substance, we will use Einstein's mass-energy equivalence principle, given by the famous equation:
\(E = mc^2\)
where:
- \(E\) is the energy equivalent.
- \(m\) is the mass of the substance.
- \(c\) is the speed of light, approximately \(3 \times 10^8 \, \text{m/s}\).
For 1 gram of a substance:
- \(m = 1 \, \text{g} = 1 \times 10^{-3} \, \text{kg}\) (since 1 gram is \(0.001\) kilograms)
Substituting these values into the equation:
\(E = (1 \times 10^{-3} \, \text{kg}) \times (3 \times 10^8 \, \text{m/s})^2\)
Calculate the value inside the parentheses:
\((3 \times 10^8 \, \text{m/s})^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2\)
Now, multiply them together:
\(E = (1 \times 10^{-3}) \times (9 \times 10^{16})\)
\(E = 9 \times 10^{13} \, \text{joules}\)
To convert joules to electronvolts (eV), use the conversion factor:
\(1 \, \text{joule} = 6.242 \times 10^{18} \, \text{eV}\)
Therefore,
\(E = 9 \times 10^{13} \, \text{joules} \times 6.242 \times 10^{18} \, \text{eV/joule}\)
\(E = 5.6178 \times 10^{32} \, \text{eV}\)
Now, convert electronvolts to mega-electronvolts (MeV):
\(1 \, \text{MeV} = 10^6 \, \text{eV}\)
\(E = \frac{5.6178 \times 10^{32} \, \text{eV}}{10^6}\)
\(E = 5.6178 \times 10^{26} \, \text{MeV}\)
Thus, the energy equivalent of 1 g of substance is \(5.6 \times 10^{26} \, \text{MeV}\), which matches the given correct option.
Approach Solution -2
The energy equivalent of a mass is given by Einstein’s equation:
\( E = mc^2, \)
where:
- \( m = 1 \, \text{g} = 10^{-3} \, \text{kg}, \)
- \( c = 3 \times 10^8 \, \text{m/s}. \)
Step 1: Substitute values into \( E = mc^2 \)
\( E = (10^{-3}) \times (3 \times 10^8)^2 \, \text{J}. \)
Simplify:
\( E = (10^{-3}) \times (9 \times 10^{16}) \, \text{J}. \)
\( E = 9 \times 10^{13} \, \text{J}. \)
Step 2: Convert energy to electron volts (eV)
Using the conversion \( 1 \, \text{J} = 6.241 \times 10^{18} \, \text{eV}: \)
\( E = (9 \times 10^{13}) \times (6.241 \times 10^{18}) \, \text{eV}. \)
\( E = 56.169 \times 10^{31} \, \text{eV}. \)
Convert to MeV (\( 1 \, \text{MeV} = 10^6 \, \text{eV} \)):
\( E = 56.169 \times 10^{25} \, \text{MeV}. \)
Approximate:
\( E \approx 5.6 \times 10^{26} \, \text{MeV}. \)
Final Answer: \( 5.6 \times 10^{26} \, \text{MeV}. \)
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Modern Physics Questions
- A long straight wire of circular cross-section (radius \( a \)) is carrying a steady current \( I \). The current \( I \) is uniformly distributed across this cross-section. The magnetic field is:
- Two sources of light emit with a power of 200 W.The ratio of number of photons of visible light emitted by each source having wavelengths 300 nm and 500 nm respectively, will be :
- A hydrogen atom is bombarded with electrons accelerated through a potential difference of \( V \), which causes excitation of hydrogen atoms. If the experiment is being performed at \( T = 0 \, \text{K} \), the minimum potential difference needed to observe any Balmer series lines in the emission spectra will be\[\frac{\alpha}{10} V,\]where \( \alpha = \, \underline{\hspace{2cm}} \).
- From the statements given below :
(A) The angular momentum of an electron in nth orbit is an integral multiple of h.
(B) Nuclear forces do not obey inverse square law.
(C) Nuclear forces are spin dependent.
(D) Nuclear forces are central and charge independent.
(E) Stability of nucleus is inversely proportional to the value of packing fraction.
Choose the correct answer from the options given below : - Work function for an object is 2.3 eV. If maximum kinetic energy of ejected electrons is 0.18 eV, find wavelength \( \lambda \) of incident photon on object.
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.