00:01
So maximum wavelength to emit the electron from the titanium metal is given.
00:06
So to emit electrons from titanium metal, the maximum wavelength required is given that is 286 nanometer.
00:26
And we have to find out the minimum energy of the photons to emit the electrons from titanium metal via photoelect.
00:34
Electric effect.
00:36
So we know that the minimum energy required to emit electrons is equals to hc by lambda.
00:58
So it is equal to hc by lambda maximum.
01:03
And here in this formula, h is plant's constant.
01:12
H is the plant's constant, which is equals to 6 .626 .26 into 10 ,000, to the power minus 34 jule into second and c is the speed of light speed of light which is equals to 3 into 10 to the power 8 meter per second and here the lambda maximum is the wavelength of light which is given as 286 nanometer so now we can put the values and find out the energy corresponds to the lambda value 286 nanometer.
02:02
So now we'll put all those values and find the energy.
02:06
So energy will be equals to 6 .626 into 10 to the power minus 34 into 3 into that is speed of light 10 to the power 8 divided by lambda maximum which is given as 286 nanometer.
02:31
So if we'll convert it to meter then it will be 286 in.
02:34
To 10 to the per minus 9 meter.
02:37
So after calculating it will get 6 .95 multiplied by 10 to the per minus 19 jr.
02:47
So the energy corresponds to the lambda value 286.
02:53
95 into 10 to the power minus 19 jr.
02:59
So this is the energy corresponds to the lambda value 286 nanometer.
03:05
So so now the second question is we have to find out the frequency of the radiation.
03:14
So frequency we know that the frequency of radiation.
03:22
So frequency of radiation is mu and we know that mu is equal to c by lambda.
03:36
And here c is the speed of light...