A surface has a work function 1.70 eV and has light of wavelength 620 nm shone o

Question

A surface has a work function 1.70 eV and has light of wavelength 620 nm shone on it. What is the kinetic energy of the photoelectrons emitted? 0.30 eV 1.40 eV 1.70 eV 2.00 eV 3.00 eV If the momentum of an electron is 1.95 times 10^-27 ke m is what is its de Broglie wavelength? 340 nm 210 nm 170 nm 420 nm 520 nm What is the de Broglie wavelength of the electron whose kinetic energy is 1.0 eV? 520 nm 8.2 nm 0.64 nm 52 nm 1.2 nm If the accuracy in measuring the position of a particle increases, the accuracy in measuring its velocity will increase. decrease remain the same. be uncertain The figure below shows the energy level diagram of a hypothetical atom Light in the visible portion of the spectrum has a wavelength between 400 nm and 750 nm. What is the maximum number of spectral line for this atom? 1 3 4 6 How many lines are possible in the visible spectrum? 0 1 2 3 What is the energy required to ionize an atom of this type that is originally in its ground state? -6.0 eV 1.0 eV 2.5 eV 6.0 eV 13.6 eV What is the wavelength of light corresponding to a transition from the a - 3 to 2 state? 1240 nm 656 nm 122 nm 827 nm

Explanation / Answer

8.

KE =hc/lambda -W

KE =(6.626*10-34)(3*108)/(620*10-9)(1.6*10-19)-1.7

KE=0.3 eV

9.

Debroglie wavelength

lambda=h/p =(6.626*10-34)/(1.95*10-27)

lambda =340 nm

10.

Kinetic energy

KE=(1/2)mV2

1.6*10-19 =(1/2)*(9.11*10-31)*V2

V=5.92*105 m/s

momentum

p=mV =(9.11*10-31)(5.92*105) = 5.4*10-25

wavelength

lambda =h/p =6.626*10-34/(5.4*10-25) =1.2 nm

11.

decreases

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