Question

***Please help fill in the table...finding KE and Frequency for each

Frequency (Hz) Metal Wavelength (nm) Stopping Voltage (V) KE f = i Sodium 247 2.68 Zinc 162 3.30 Copper 152 3.39 Platinum 122 3.79 Calcium 213 2.86

Answer 1

Follow table for short :
1- = 8.68 X 1.6x1079 "KE is given by : (stopping voltage) x (electronic charge) KE =) ev ve=1-6 X 10-19 Hence ① For Na V.2.68V KIE Similarly for Zinc Ve 3.30 V KE - 3.30 X 1-68 X1079 5.894 x 10^4J For cu v=3.39 kita 3-39 X 1.6 x 1019 5.424 x 1079 ] For Platinum v.3.79v Kie : 3.79 x 16 x 1079 6.064 x 1079 6 for Calcium! V=2.86 N kie s.2-86 81.6X 1019 & 14.576 x 10193 → Now Frequency in d = c/a For Na! d=247 nm speed of 3x108 light = 3x 108 m/s à El. 21 x 105 HZ given by 247x10-9

o 15 Hz 8 For an = 162 nm da 3x 108 11.85 x 10's 162X 109 cu! d = 152 nm 3) 1 - 3x10 1.97X 10'5 HZ 1528109 for pt ! t= 122 nm ta = 12.45 8 10 15 Hz (22 x long s ) 3 x 108 for ca Į d=213 nm 3x 108 1.40 x 1015 Hz 213 x 10g Metal Wavelength (on) stopping K.E(J) J =%A potential (v) Cha Sodium 247 2.68 4.288 X 1049 1.21x10ls 1.858 1015 3.30 162 Zn 5-288 1019 152 3.39 8-424 X 1019 1-94X1005 Pt 122 3.79 6.064 x 1019 2.488lo's Calcium 213 2-86 4.576x1019 1.40 x 1015