Ionization energy of gaseous Na atoms is 495.5 kJ mol-1 The lowest possible frequency of light that ionizes a sodium atom is (h = 6.626x10-34 Js, Na = 6.022 × 1023 mol-1)

Question

Ionization energy of gaseous Na atoms is 495.5 kJ mol-1 The lowest possible frequency of light that ionizes a sodium atom is (h = 6.626x10-34 Js, Na = 6.022 × 1023 mol-1) (A) 7.50 × 104 s-1 (B) 4.76 × 1014 s-1 (C) 3.15 × 1015 s-1 (D) 1.24 × 1015 s-1

Tardigrade Admin · Accepted Answer

Δ E = hv v =(Δ E / h ) v =(495.5 × 103 Joule /6.023 × 1023) × (1/6.626 × 10-34) v =1.24 × 1015 sec -1

Q. Ionization energy of gaseous Na atoms is $495.5 k J m o l^{- 1}$ The lowest possible frequency of light that ionizes a sodium atom is (h = 6.626x10^-34 Js, Na = 6.022 $\times$ 1023 mol^-1)

$7.50 \times 1 0^{4} s^{- 1}$

$4.76 \times 1 0^{14} s - 1$

$3.15 \times 1 0^{15} s - 1$

$1.24 \times 1 0^{15} s - 1$

$Δ E = h v$

$v = \frac{Δ E}{h}$

$v = \frac{495.5 \times 1 0 ^{3} J o u l e}{6.023 \times 1 0 ^{23}} \times \frac{1}{6.626 \times 1 0 ^{- 34}}$

$v = 1.24 \times 1 0^{15} se c^{- 1}$

Q. Ionization energy of gaseous Na atoms is 495.5kJmol−1 The lowest possible frequency of light that ionizes a sodium atom is (h = 6.626x10-34 Js, Na = 6.022 × 1023 mol-1)

7.50×104s−1

4.76×1014s−1

3.15×1015s−1

1.24×1015s−1