A hydrogen atom in its ground state absorbs 10.2 eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of: (Given, Planck’s constant = \(6.6 × 10^{–34}\) Js).
\(2.10 × 10^{–34}\) Js
- \(1.05 × 10^{–34}\) Js
- \(3.15 × 10^{–34}\) Js
- \(4.2 × 10^{–34}\) Js
The Correct Option is B
Solution and Explanation
\(-13.6+10.2=\frac{-13.6}{n^2}\)
\(⇒ \frac{13.6}{n^2} = 3.4\)
\(⇒ n = 2\)
\(⇒ Δ L = 2 \times \frac{h}{2λ} - 1 \times \frac{h}{2λ}\)
\(⇒ \frac{h}{2λ}\)
\(⇒ Δ L ≅ 1.05 \times 10^{-34} \;J\;s\)
Hence, the correct option is (B): \(1.05 × 10^{–34}\) \(J s\)
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Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Knowing the initial position \( x_0 \) and initial momentum \( p_0 \) is enough to determine the position and momentum at any time \( t \) for a simple harmonic motion with a given angular frequency \( \omega \).
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Questions Asked in JEE Main exam
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Let one focus of the hyperbola \( H : \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 \) be at \( (\sqrt{10}, 0) \) and the corresponding directrix be \( x = \dfrac{9}{\sqrt{10}} \). If \( e \) and \( l \) respectively are the eccentricity and the length of the latus rectum of \( H \), then \( 9 \left(e^2 + l \right) \) is equal to:
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Concepts Used:
Momentum
It can be defined as "mass in motion." All objects have mass; so if an object is moving, then it is called as momentum.
the momentum of an object is the product of mass of the object and the velocity of the object.
Momentum = mass • velocity
The above equation can be rewritten as
p = m • v
where m is the mass and v is the velocity.
Momentum is a vector quantity and the direction of the of the vector is the same as the direction that an object.