Radiation of wavelength 242 nm is sufficient to ionise the sodium atom. Calculate the ionization energy in $k J m o l^{- 1}$ . ( $h = 6.625 \times 10^{- 34} J - s$ )

Question

Radiation of wavelength 242 nm is sufficient to ionise the sodium atom. Calculate the ionization energy in

k J m o l^{- 1}

. (

h = 6.625 \times 10^{- 34} J - s

)

Toppr Admin · Accepted Answer

IE-hc-x3BB-6-625-xD7-10-x2212-34-xD7-3-xD7-108242-xD7-10-x2212-9-xD7-NA-xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0-494657 joule-mol-xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0- -xA0-494-65 Kj-mol

Radiation of wavelength $242$ nm is sufficient to ionize the sodium atom. Calculate the ionization of energy in kJ mol $^{- 1}$ . ( $n = 6.625 \times 10^{- 24}$ J-s).
$494.57 K J m o l^{- 1}$
$594.57 K J m o l^{- 1}$
$694.57 K J m o l^{- 1}$
$394.57 K J m o l^{- 1}$

$I E = \frac{h c}{λ} = (\frac{6.625 \times 10^{- 34} \times 3 \times 10^{8}}{242 \times 10^{- 9}} \times N_{A})$
$= 494657$ joule/mol
$= 494.65$ Kj/mol

Radiation of wavelength 242 nm is sufficient to ionize the sodium atom. Calculate the ionization of energy in kJ mol−1. (n=6.625×10−24J-s).494.57 KJmol−1594.57 KJmol−1694.57 KJmol−1394.57 KJmol−1

IE=hcλ=(6.625×10−34×3×108242×10−9×NA) =494657 joule/mol =494.65 Kj/mol

Radiation of wavelength $242$ nm is sufficient to ionize the sodium atom. Calculate the ionization of energy in kJ mol $^{- 1}$ . ( $n = 6.625 \times 10^{- 24}$ J-s).
$494.57 K J m o l^{- 1}$
$594.57 K J m o l^{- 1}$
$694.57 K J m o l^{- 1}$
$394.57 K J m o l^{- 1}$

$I E = \frac{h c}{λ} = (\frac{6.625 \times 10^{- 34} \times 3 \times 10^{8}}{242 \times 10^{- 9}} \times N_{A})$
$= 494657$ joule/mol
$= 494.65$ Kj/mol