4000 A photon is used to break the iodine molecule, then the % of energy converted to the K.E. of iodine atoms if bond dissociation energy of I₂ molecule is 246.5 kJ/mol

        (A)    8%

        (B)    12%

        (C)    17%

        (D)    25%

Question

4000 A photon is used to break the iodine molecule, then the % of energy converted to the K.E. of iodine atoms if bond dissociation energy of I₂ molecule is 246.5 kJ/mol

(A) 8%

(B) 12%

(C) 17%

(D) 25%

Aditi Chauhan · Answer

(C)

Sol. Energy of one photon =12400/4000

= 3.1 ev

Energy supplied by one ole photon in KJ/mole = 3.1 × 1.6 × 10^–19 × 6 × 10²³ × 10^–3 = 297 kJmol^–1

% of energy converted to K.E. = = 17%

Krishi Patel · Answer

Calculate the work function for metal for which ejection of the photoelectron can be stopped by applying 0.4 V when the radiation of 300nm is used.

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4000 A photon is used to break the iodine molecule, then the % of energy converted to the K.E. of iodine atoms if bond dissociation energy of I 2 molecule is 246.5 kJ/mol (A) 8% (B) 12% (C) 17% (D) 25%

4000 A photon is used to break the iodine molecule, then the % of energy converted to the K.E. of iodine atoms if bond dissociation energy of I₂ molecule is 246.5 kJ/mol

        (A)    8%

        (B)    12%

        (C)    17%

        (D)    25%

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