Match the following:

A particular station of All India Radio, New Delhi, broadcasts on a frequency of 1,368 kHz (kilohertz). The wavelength of the electromagnetic radiation emitted by the transmitter is: [speed of light, $\mathrm{c}=3.0\times {10}^8$ ${\mathrm{ms}}^{-1}$]

A yellow radiation having wavelength 5800 Å. The wavenumber (a) and frequency (b) of yellow radiation are respectively-
1. \(\mathrm{a   = 172.4   \times   10^{4}    ~ cm^{- 1} ,   b   =   5 . 172   \times   10^{14} ~  s^{- 1}}\)
2. \(\mathrm{a   =   5 . 172   \times   10^{14} ~  s^{- 1} ,   b   = 17.24   \times   10^{4}   ~  cm^{- 1}}\)
3. \( \mathrm{a   =   724   \times  10^{5}  ~   cm^{- 1} ,   b   =   5 . 172   \times   10^{15} ~  s^{- 1}}\)
4. \(\mathrm{a   = 1.72 4   \times   10^{4}   ~ cm^{- 1} ,   b   =   5 . 172   \times   10^{14} ~ s^{- 1}}\)

A monochromatic infrared range finder of power 1milli watt emits photons with wavelength 1000 nm in 0.1 second. The number of photons emitted in 0.1 second is:
(Given: h = \(6.626 \times 10^{-34} J~s\) , c = \(3 \times 10^8 m~s^{-1}, \) Avogadro number = \(6.022 \times 10^{23}\))

1. \(30 \times 10^{37}\)
2. \(5 \times 10^{14} \)
3. \(30 \times 10^{34} \)
4. \(5 \times 10^{11} \)

The energy of the photon with a frequency of 3 × 10¹⁵ Hz is:

1. 1.988 × ${10}^{-18}$ J
2. 1.588 × ${10}^{-18}$ J
3. 1.288 × ${10}^{-18}$ J
4. 2.988 × ${10}^{-18}$ J

The correct arrangement of the following electromagnetic spectrum in the increasing order of frequency is:

The neon gas emits radiation of 616 nm. The number of quanta that are present in 2 J of energy is:

1. \(6.2 \times 10^{-18} \)
2. \(5.6 \times 10^{17} \)
3. \(6.2 \times 10^{18} \)
4. \(32.2 \times 10^{-20}\)