The total number and mass of neutrons in 7 mg of ¹⁴C would be :

(Assume that mass of a neutron = 1.675 × 10^–27 kg)

\(1 .\) \(2 . 41\) \(\times\) \(\left(10\right)^{21}\) \(,\) \(4 . 03\) \(\times\) \(\left(10\right)^{– 6}\) \(kg\)
\(2 .\) \(6 . 23\) \(\times\) \(\left(10\right)^{23}\) \(,\) \(1 . 67\) \(\times\) \(\left(10\right)^{- 21} kg\)
\(3 .\) \(1 . 22\) \(\times\) \(\left(10\right)^{22}\) \(,\) \(4 . 03\) \(\times\) \(\left(10\right)^{6}\) \(kg\)
\(4 .  2 . 41\) \(\times\) \(\left(10\right)^{21}\) \(,\) \(4 . 03\) \(\times\) \(\left(10\right)^{- 6}\) \(g\)

The wave number of a light whose time period is 2.0 × 10^–10 s would be:

1. 16.66 m^-1

2. 1.66 m^-1

3. 32.34 m^-1

4. 12.34 m^-1

The number of photons of light with a wavelength of 4000 pm that provide 1J of energy would be: 

$1.$ $2.01$ $\times$ ${10}^{16}$
$2.$ $2.01$ $\times$ ${10}^{19}$
$3.$ $4.14$ $\times$ ${10}^{23}$
$4.$ $2.14$ $\times$ ${10}^{21}$

Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise sodium atom. The ionisation energy of sodium in kJ mol^-1 is -

1. 494

2. 4.94

3. 516

4. 0.50

A 25-watt bulb emits monochromatic yellow light with a wave length of 0.57µm. The rate of emission of quanta per second would be :

1. 7.17×10^-19 s^-1

2. 4.13×10¹⁶s-¹

3 . 7.17×10¹⁹ s^-1

4 . 1.26 ×10²⁰s^-1
${\mathrm{}}^{}$

The wavelength of light emitted when the electron in a H atom undergoes the transition from an energy level with n = 4 to an energy level with n = 2, is :

1. 586 mm

2. 486 nm

3. 523 nm

4. 416 pm

The energy associated with the fifth orbit of a hydrogen atom is :

$1.$ $-2.18$ $\times$ ${10}^{-18}$ $\mathrm{J}$
$2.$ $-8.72$ $\times$ ${10}^{-20}$ $\mathrm{J}$
$3.$ $-3.88$ $\times$ ${10}^{-21}$ $\mathrm{J}$
$4.$ $-8.72$ $\times$ ${10}^{-19}$ $\mathrm{J}$

The wave number for the longest wavelength transition in the Balmer series of atomic hydrogen would be -
1. \(1 .   1 . 52   \times   \left(10\right)^{6}   m^{- 1}\)
2. \(2 .   3 . 14   \times   \left(10\right)^{6}   \left(cm\right)^{- 1}\)
3. \(3 .   15 . 2   \times   \left(10\right)^{6}   m^{- 1}\)
4. \(4 .   1 . 52   \times   \left(10\right)^{6}   \left(cm\right)^{- 1}\)

The energy of an electron in an H - atom is given by \(E_n=(-2.18 \times10^{-18})/n^2~\mathrm J.\) The shortest wavelength of light that can be used to remove an electron completely from \(n = 2\) orbit will be:

The wavelength of an electron moving with a velocity of 2.05 × 10⁷ m s^-1  would be:  

\(1 .\) \(4 . 65\) \(\times\) \(\left(10\right)^{- 12}\) \(m\)

\(2 .\) \(3 . 55\) \(\times\) \(\left(10\right)^{-11}\) \(m\)

\(3 .\) \(2 . 34\) \(\times\) \(\left(10\right)^{11}\) \(m\)

\(4 .\) \(6 . 43\) \(\times\) \(\left(10\right)^{ -11}\) \(m\)