# A photon of light with wavelength 700 nm has energy of photon of light of what wavelength would correspond to energy 2E

Umakant biswal
5349 Points
7 years ago
@ sakshi
the question is asking about photons , so, we can directly apply the following formula ..
E= Hv here h is the planks constant , and v is the frequency .
and as u knows v = c/ lambda
where c is the speed of light and lambda is the wavelength .
so, by plugging in the value of c in equation u will get
E= HC/LAMBDA .
now put the value of lambda and calculate E ,
E= 6.62*10^-34 *3*10^8 / 700*10^-9
CALCULATE THE ABOVE AND U WILL GET E FROM THEIR .
NOW, COMES TO THE 2ND PART OF QUESTION ,
ITS ASKING ABOUT CALCULATION OF LAMBDA WHEN 2E IS GIVEN
SO, FORMULATE THE EQUATION
2E = 2 HC/ LAMBDA
AND FROM THE FIRST EQUATION U HAVE GOT THE VALUE OF E ,
MULTIPLY IT BY 2 , AND APPLY THAT HERE ,
U KNOWS THE REST THING
h= 6.62*10^-34
c= 3*10^8
and lambda u have to calculate ..
put the below formula
LAMBDA = HC/E
AND CALCULATE IT ,
hope , u can calculate it now ,
IF NOT ABLE TO GET THE RIGHT ANSWER , THEN LET ME KNOW ..
ALL THE BEST ..