A) When the tube is filled with mercury vapor, as in this case, a sharp drop in

Question

A) When the tube is filled with mercury vapor, as in this case, a sharp drop in the collected current is observed when the accelerating potential reaches 4.9 . The current drops suddenly to a negligible value and slowly increases again when the accelerating potential is above 4.9 . What is the energy absorbed by the atomic electrons in the mercury atom when the accelerating potential is 4.9 ?
B) The excited mercury atoms decay back to the lowest energy level by emitting radiation. What is the wavelength of this radiation?

Explanation / Answer

(A)
The loss in kinetic energy = gain in potentialenergy                                     E = eV                                         =1.6 x 10-19 * 4.9                                        = 7.84 x 10-19 J                                        = 7.84 x 10-19 x 6.242 x 1018                                        = 4.89 eV ------------------------------------------------------------------------------------ (B) we have E = hc / or
Wave length = hc / E here     E = 4.89 eV Thus      = (6.63*10-34J.s)(3*108m/s) / (4.89 eV)(1.6*10-19J/1eV)           =253.69*10-9m           =253.69 nm . ------------------------------------------------------------------------------------ (B) we have E = hc / or
Wave length = hc / E here     E = 4.89 eV Thus      = (6.63*10-34J.s)(3*108m/s) / (4.89 eV)(1.6*10-19J/1eV)           =253.69*10-9m           =253.69 nm . we have E = hc / or
Wave length = hc / E here     E = 4.89 eV Thus      = (6.63*10-34J.s)(3*108m/s) / (4.89 eV)(1.6*10-19J/1eV)           =253.69*10-9m           =253.69 nm . we have E = hc / or
Wave length = hc / E here     E = 4.89 eV Thus      = (6.63*10-34J.s)(3*108m/s) / (4.89 eV)(1.6*10-19J/1eV)           =253.69*10-9m           =253.69 nm .

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