JEE Main 2014 · 19 Apr · Shift 1 · Chemistry Q31

← JEE Main 2014 · 19 Apr · Shift 1

ChemistrySingle correct+4 / -1Question 31 of 90

Ionization energy of gaseous Na atoms is 495.5 kJmol $^{-1}$ . The lowest possible frequency of light that ionizes a sodium atom is ( $h=6.626\times10^{-34}$ Js, $N_A=6.022\times10^{23}$ mol $^{-1}$ )

7.50\times10^4

^{-1}

4.76\times10^{14}

^{-1}

3.15\times10^{15}

^{-1}

1.24\times10^{15}

^{-1}

Answer and worked solution

Official answer: D

Energy per atom: $\dfrac{495.5\times10^3}{6.022\times10^{23}}=8.23\times10^{-19}$ J.
$\nu=\frac Eh=\frac{8.23\times10^{-19}}{6.626\times10^{-34}}\approx1.24\times10^{15}\text{ s}^{-1}.$

Topics:Photoelectric effect and spectra Ionisation enthalpy

← Previous question Next question →

JEE Main 2014 · 19 Apr · Shift 1, Chemistry question 31