Question

DIRECTIONS: Answer the following problems, clearly showing your work below and your answer in the spaces provided on the answer sheet. Use the appropriate units and sigfigs. 51. [2 pt] The work function of gold is 8.49 x $10^{-19}$ J. What frequency (in s?¹) of light is required to eject an electron from gold such that the electron has a velocity of 100.0 m/s? 52. [3 pt] Compute the first three absorption energies (in J) of the hydrogen atom. Report your answers with 3 sigfigs. (HINT: n = 1 to n = 2, 3, and 4.) 53. [2 pt] The dipole moment of CaCl is 4.257 D and the bond length is 2.439 Å. Compute the magnitude of the charge separation (Q, in partial electrons) for CaCl.

          DIRECTIONS: Answer the following problems, clearly showing your work below and your answer in
the spaces provided on the answer sheet. Use the appropriate units and sigfigs.
51. [2 pt] The work function of gold is 8.49 x $10^{-19}$ J. What frequency (in s?¹) of light is required to
eject an electron from gold such that the electron has a velocity of 100.0 m/s?
52. [3 pt] Compute the first three absorption energies (in J) of the hydrogen atom. Report your
answers with 3 sigfigs. (HINT: n = 1 to n = 2, 3, and 4.)
53. [2 pt] The dipole moment of CaCl is 4.257 D and the bond length is 2.439 Å. Compute the
magnitude of the charge separation (Q, in partial electrons) for CaCl.

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DIRECTIONS: Answer the following problems, clearly showing your work below and your answer in
the spaces provided on the answer sheet. Use the appropriate units and sigfigs.
51. [2 pt] The work function of gold is 8.49 x 10^-19 J. What frequency (in s?¹) of light is required to
eject an electron from gold such that the electron has a velocity of 100.0 m/s?
52. [3 pt] Compute the first three absorption energies (in J) of the hydrogen atom. Report your
answers with 3 sigfigs. (HINT: n = 1 to n = 2, 3, and 4.)
53. [2 pt] The dipole moment of CaCl is 4.257 D and the bond length is 2.439 Å. Compute the
magnitude of the charge separation (Q, in partial electrons) for CaCl.

Added by Juan Carlos A.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

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Solved by Expert Adi S

07/18/2022

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626 x 10^-34 J*s) f = frequency of the light φ = work function of gold Show more…

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51. The work function of gold is 8.49 x 10^-19 J. What frequency (in s^-1) of light is required to eject an electron from gold such that the electron has a velocity of 100.0 m/s? 52. Compute the first three absorption energies (in J) of the hydrogen atom. Report your answers with 3 sig figs. (HINT: n = 1 to n = 2, 3, and 4.) 53. The dipole moment of CaCl is 4.257 D and the bond length is 2.439 Å. Compute the magnitude of the charge separation (Q, in partial electrons) for CaCl.

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Transcript

00:01 Work function for the goal is given in the question as q is equal to 5 .10 electron volts.

00:11 In the first equation, they ask to find the work function of the gold from electron volts to joules.

00:20 So q is equal to 5 .10 electron volts.

00:25 One electron value 1 .6 into 10 to the power of minus 19 joules.

00:31 So by multiplying this we got 3 .16 into 10 to the power of minus 90 joules.

00:42 The second question, they asked to find the cutoff frequency.

00:51 For cutoff frequency, charge is equal to n f.

00:55 When n is the number of electrons, f is the frequency.

00:59 So that implies cutoff frequency is equal to charge by number of electrons.

01:08 Charge already we calculated that 3 .16 into 10 to the 10 to the power of minus 19 joules by number of electrons in one unit of charge that is 6 .625 into 10 to the power of minus 34 so by doing the calculation we got 1 .23 into 10 to the power of 15 heads.

01:42 So this is the cutoff frequency of the...

01:47 In the third question, they ask to calculate the maximum wavelength of light incident on a goal to release the photo electrons from the surface.

01:58 For that sake, the frequency is equal to...

02:06 Cutoff frequency is equal to speed of light divided by wavelength and that cutoff frequency already we calculated that is 1 .2 3 into 10 to the power of 15 is equal to see that is this light speed 3 into 10 to the power of 8 meter per second divided by lambda nothing but a wavelength...

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