/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 84 The energy, the magnitude of lin... [FREE SOLUTION] | 91Ó°ÊÓ

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The energy, the magnitude of linear momentum and orbital radius of an electron in a hydrogen atom corresponding to the quantum number \(n\) are \(E, P\) and \(r\) respectively. Then according to Bohr's theory for hydrogen atom (A) \(E P r\) is proportional to \(\frac{1}{n}\). (B) \(\frac{P}{E}\) is proportional to \(n\). (C) \(E r\) is constant for all orbits. (D) \(P r\) is proportional to \(n\).

Short Answer

Expert verified
According to Bohr's theory for hydrogen atom, the correct statements are: (C) \(Er\) is constant for all orbits. (D) \(Pr\) is proportional to \(n\).

Step by step solution

01

Find the expression for energy (E)

According to Bohr's theory, the energy of an electron in a hydrogen atom is given by: \(E = -\frac{13.6 eV}{n^2}\) Where n is the principal quantum number (n = 1, 2, 3, ...)
02

Find the expression for linear momentum magnitude (P)

The linear momentum magnitude (P) can be calculated using the de Broglie wavelength: \(P = \frac{h}{\lambda}\) The circumference of the orbit is an integral multiple of the de Broglie wavelength: \(2 \pi r = n\lambda \) Combining the two equations, we get: \(P = \frac{h}{\lambda} = \frac{hn}{2 \pi r}\)
03

Find the expression for orbital radius (r)

The expression for the orbital radius in terms of the principal quantum number (n) is given by: \(r = \frac{n^2h^2}{4\pi^2me^2}\) Where \(h\) is Planck's constant, \(m\) is the mass of the electron, and \(e\) is the elementary charge.
04

Examine the given statements

(A) Is \(EPr\) proportional to \(\frac{1}{n}\)? Using the derived expressions for \(E\), \(P\), and \(r\): \(EPr = \left(-\frac{13.6}{n^2}\right) \left(\frac{hn}{2\pi r}\right) \left(\frac{n^2h^2}{4\pi^2me^2}\right)\) Simplifying, all the terms containing \(n\) cancel out. Thus, \(EPr\) is not proportional to \(\frac{1}{n}\), and Statement A is False. (B) Is \(\frac{P}{E}\) proportional to \(n\)? Using the expressions for \(E\) and \(P\): \(\frac{P}{E} = \frac{\frac{hn}{2\pi r}}{-\frac{13.6}{n^2}}\) After plugging in the expression for \(r\), we find that all the terms containing \(n\) cancel out. Thus, \(\frac{P}{E}\) is not proportional to \(n\), and Statement B is False. (C) Is \(Er\) constant for all orbits? Using the derived expressions for \(E\) and \(r\): \(Er = \left(-\frac{13.6}{n^2}\right)\left(\frac{n^2h^2}{4\pi^2me^2}\right)\) Simplifying, all the terms containing \(n\) cancel out, resulting in a constant value. Thus, statement C is true. (D) Is \(Pr\) proportional to \(n\)? Using the derived expressions for \(P\) and \(r\): \(Pr = \left(\frac{hn}{2\pi r}\right)\left(\frac{n^2h^2}{4\pi^2me^2}\right)\) Simplifying, we find that \(Pr\) is proportional to \(n\). Thus, statement D is true.
05

Conclusion

According to Bohr's theory for hydrogen atom: (A) False: \(EPr\) is not proportional to \(\frac{1}{n}\) (B) False: \(\frac{P}{E}\) is not proportional to \(n\) (C) True: \(Er\) is constant for all orbits (D) True: \(Pr\) is proportional to \(n\)

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