The kinetic energy of emitted electrons when cesium is exposed to UV rays of frequency 1.80 * 10 ^15 is 3.054 x 10 ^- 19 J.
Explanation:
To calculate the kinetic energy of emitted electrons,
It is given that the frequency is 1.80* 10^15 Hz
We have,
KE = E – Eo = hv –hvo
Where, h = 6.626 x 10^ - 34 Js
Given frequency = 1.4 x 10 ^ 15 Hz
vo (Threshold frequency) for cesium = 9.39 x 10 ^ 14 Hz
Applying in equation,
we get
KE = 6.626 x 10^34 (1.4 x 10^15 - 9.39 x 10^14)
KE= 3.054 x 10^-19 J
[Note: Here, threshold frequency of Cesium is not provided. Apply the correct threshold frequency from the part A]
Answer:
Answer is 5.71 *10^-19 J
Explanation:
KE = E – Eo = hv –hvo
h= 6.63*10^-34 J*s is planck's constant
v= 1.8*10^15 HZ is given frequency
v0= 9.39*10^14
When you plug in the values to the equation above, you get:
KE= (1.1934*10^-18) - (6.22557*10^-19)
KE=5.70843*10^-19
Now to 3 sig figs
KE=5.71*10^-19
