What is the kinetic energy of the emitted electrons when cesium is exposed to uv rays of frequency 1.80×1015hz ?

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]

the answer is minerals.

the dissociation of nickel hydroxide is follows:

the concentration of hydroxide ion is

the equilibrium constant value of nickel hydroxide is = 6.1 * 10^-^1^6 and thus equilibrium equation can be written as:

thus, concentration of ni is 0.23 * 10^-^4 mol /l