JEE Main Atom PYQ

36. (AIEEE 2008)

Suppose an electron is attracted towards the origin by a force $\frac{k}{r}$ where $^{'} k^{'}$ is a constant and $^{'} r^{'}$ is the distance of the electron from the origin. By applying Bohr model to this system, the radius of the $n^{t h}$ orbital of the electron is found to be $^{'} {r_{n}}^{'}$ and the kinetic energy of the electron to be $^{'} {T_{n}}^{'} .$

Then which of the following is true?

A. $T_{n} \propto \frac{1}{n^{2}}, r_{n} \propto n^{2}$

B. $T_{n}$ independent of $n, r_{n} \propto n$

C. $T_{n} \propto \frac{1}{n}, r_{n} \propto n$

D. $T_{n} \propto \frac{1}{n}, r_{n} \propto n^{2}$

The correct answer is option (B)

When $F = \frac{k}{r} =$ centripetal force, then $\frac{k}{r} = \frac{m v^{2}}{r}$

$\Rightarrow m v^{2} =$ constant $\Rightarrow$ kinetic energy is constant

$\Rightarrow T$ is independent of $n .$

37. ( AIEEE 2003)

Which of the following atoms has the lowest ionization potential ?

A. $_{7}^{14} N$

B. $_{55}^{133} C s$

C. $_{18}^{40} A r$

D. $_{8}^{16} O$

The correct answer is option (C)

The ionisation potential increases from left to right in a period and decreases from top to bottom in a group. Therefore ceasium will have the lowest ionisation potential.

38. (AIEEE 2003)

If the binding energy of the electron in a hydrogen atom is $13.6 e V,$ the energy required to remove the electron from the first excited state of $L i^{+ +}$ is

A. $30.6$ $e V$

B. $13.6$ $e V$

C. $3.4$ $e V$

D. $122.4$ $e V$

The correct answer is option (C)

$E_{n} = - \frac{13.6}{n^{2}} Z^{2} e V /$ atom

For lithium ion $Z = 3;$ $n = 2$ (for first excited state)

$E_{n} = - \frac{13.6}{2^{2}} \times 3^{2} = - 30.6 e V$

39. (AIEEE 2002)

If $13.6$ $e V$ energy is required to ionize the hydrogen atom, then the energy required to remove an electron from $n = 2$ is

A. $10.2$ $e V$

B. $0$ $e V$

C. $3.4$ $e V$

D. $6.8$ $e V .$

The correct answer is option (C)

KEY CONCEPT :

The energy of nth orbit of hydrogen is given by

$E_{n} = \frac{13.6}{n^{2}} e V /$ atom

For $n = 2,$ $E_{n} = \frac{- 13.6}{4} = - 3.4 e V$

Therefore the energy required to remove electron from

$n = 2$ is $+ 3.4 e V .$