sree DEBASISH DASGUPTA

শনিবার, ২৫ জানুয়ারী, ২০২৫

New link DANS-EASY

https://explore.openaire.eu/search/find?fv0=sree%20debasish%20dasgupta&f0=q&page=1

বৃহস্পতিবার, ২৬ সেপ্টেম্বর, ২০২৪

My request to all student societies

My request to all student societies is to read my 'Atomic Philosophy in the Path of Light ' essay carefully. This article contains some information that will be useful for you in the future. Mendeley's data has been published, it can be found if you search Google. Link; https://data.mendeley.com/datasets/st6fxh2mnb/1

Below are some examples.

· Crucial Number in Particle Creation: The number '0.413334' appears to play a vital role in the creation of various particles and elements, suggesting that there may be a foundational ratio or constant that governs their formation.

Visible Light Energy and Frequency Relationship: The relationship between energy (eV) and frequency is articulated in the equation eV = 0.413334 x Frequency, suggesting a direct correlation between these two physical properties.

For Example:- Understanding the use of ‘Electron Volt = Frequency x 0.413334’ formula as unit conversions in current units calculation.

Question : (Example) - Electro light action of starting wave length is 7000 Angstrom to any metal transform it to the unit of W0 to eV .

(h=6.60 x 10^^-27 erg-s , eV = 1.6 x 10^^-12 erg)

Solve : -

We know Frequency = speed of light / wave length

So here frequency : = 4.2857 x 10^¹⁴ Hz .

As per my rule (without any power) eV = frequency x 0.413334

eV = (4.2857 x 0.413334) = 1.7714 unit (eV)

and same way frequency = eV / 0.413334

Frequency = (1.7714 / 0.413334) = 4.2857 unit (Current unit Hertz)

· Pi and Particle Masses: The constant Pi can be expressed through a complex formula that integrates the mass values of electrons and protons, illustrating a deep interconnection between these fundamental constants and the energies involved.

Pi = {[( root over electron either protons mass value energy) / (Proton energy as per e =mc²)] / 2 }^{^4}

For Example:-

[We know that one (Pi = 3.1415926535897932384626433832795 and as per “e=mc^{^2} .” 1.5032784196545335010226279575152e-10 Unit = Proton’s mass equivalent energy.

And 1.60217653e-19 Unit electric charge of a Proton then (-1.60217653e-19) Unit electric charge of an electron]

Now, 1.60217653e-19

Root over> = 4.0027197378782342122581519961685e-10

4.0027197378782342122581519961685e-10 / 1.5032784196545335010226279575152e-10 = 2.6626602800551704435263034591073 /2 = (1.3313301400275852217631517295536)

(1.3313301400275852217631517295536)^{^4} = 3.1415433470961018845922239897721 =[Pi].

Next.

( 1.60217653e-19 / 2) = 8.01088265e-20

root> 8.01088265e-20 = 2.8303502698429394217655138448226e-10

(2.8303502698429394217655138448226e-10 / 1.5032784196545335010226279575152e-10) = 1.882785140023082758292977718897

1.882785140023082758292977718897^{^4} =12.566173388384407538368895959089 = 4Pi

So, {[Root over (protons electric charge/2)]/ (Proton energy as per e =mc²)]}^{^4} =4Pi,

· Proton Mass Calculation: The mass of a proton can be derived from the electron mass using a specific formula involving the number 0.413334, thus establishing a calculative approach to understanding the mass ratio between these particles.

[(Electron mass) / ( 0.413334 x 10 )^{^2} ] x (Pi)^{^9} x [(Pi) / 3 ] = PROTON’s MASS

Pi and Particle Masses: The constant Pi can be expressed through a complex formula that integrates the mass values of electrons and protons, illustrating a deep interconnection between these fundamental constants and the energies involved.

[ (Electron mass) / (0.413334 *10) ^{^2} ] divided by (Pi)^{^9} = Mass. 1.782662e-36 unit.

Or, 1/ {(Pi)^⁹ / [(Electron mass) / (0.413334 *10)^{^2}] } = Mass, 1.782662e-36 unit.

For Example:- Investigating the impact of internal changes on the properties of particles and exploring the role of (L = 0.413334) and (Pi), in electron, protons mass and energy.

We know that , as per (e=mc^{^2}) equation. (Mass = 1.782662e-36 Kg. When energy = 1.60217653e-19 Joule.

Electron mass = 9.109390e-31kg, C^2 = speed of light square)

‘L’ = (photon’s eV 1.24 unit / 3) = 0.413334,

(0.413334 x10) ^{^2} = 17.084449955556 , either [L x (Time /space)] = [0.413334 x (3600/360)]^{^2} =17.084449955556 [3600 second & 360 Degree ]

(9.109390e-31/17.084449955556) = 5.3319773382798041975309487724255e-32

(Electron mass) / (0.413334 *10) ^{^2} = 5.3319773382798041975309487724255e-32 ---- (1a)

[1.2985270578788836 is constant unit for particle energy, and ‘L’ = 0.413334, in this article, this subject has been discussed earlier]

22/7 = (3.1428571428571428571428571428571) = Pi

(3.1428571428571428571428571428571)^{^9} = 2.9917256660402129603928590571842e+4 ---(2a)

Equation, (2a / 1a) =

(2.9917256660402129603928590571842e+4 / 5.3319773382798041975309487724255e-32)

= 5.6109121930465663243853382177961e+35

(1/5.6109121930465663243853382177961e+35) = 1.7822414e-36 Unit Mass.

As per (e=mc^{^2}) equation. (Mass = 1.782662e-36 Kg. When energy = 1.60217653e-19 J. )

Either — (1a)/(2a) =

(5.3319773382798041975309487724255e-32 / 2.9917256660402129603928590571842e+4)

. = 1.7822414e-36 Unit mass.

So [ (Electron mass) / (0.413334 *10) ^{^2} ] divided by (Pi)^{^9} = Mass.

Or, 1/ {(Pi)^⁹ / [(Electron mass) / (0.413334 *10)^{^2}] } = Mass,

[As per (e=mc^{^2}) equation. Mass = 1.782662e-36 Unit. When energy = 1.60217653e-19 Unit ]

Next , (1a)x(2a).=

(5.3319773382798041975309487724255e-32) x (2.9917256660402129603928590571842e+4)

= 1.5951813453676469100727741096844e-27.

(1.5951813453676469100727741096844e-27) * (3.1428571428571428571428571428571 /3)

= 1.6711423618137253343619538291932e-27 Unit. Proton’s mass.

So, (1a)x(2a) x [(22/7) / 3 ] or,

[(Electron mass) / ( 0.413334 x 10 )^{^2} ] x (Pi)^{^9} x [(Pi) / 3 ] = PROTON’s MASS.

We know that, as per (e=mc^{^2}) equation. Proton’s mass = 1.672623e-27 Kg.

Overall, these points underscore the intricate relationships and constants that define the behavior and characteristics of fundamental particles and the energies associated with them. They present a framework for further exploration and understanding of the principles governing atomic and subatomic phenomena.

মঙ্গলবার, ২ জুলাই, ২০২৪

My science based Articles are now available on this site as well. I am very happy.

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সোমবার, ১৬ মে, ২০২২

জয়ের পর বিজয়ীর জন্য উল্লাসে সকলে উল্লসিত হয়। কিন্তু যে পরাজিত হয়েছে তার জন্য দুঃখে কেউ কাতর হয়না

বুধবার, ৩ নভেম্বর, ২০২১

Answer from PMOPG PORTAL and PG PORTAL Govt.of india

I am compelled to post this on social media, because if the government's attention can be drawn by social media.

मैं इसे सोशल मीडिया पर पोस्ट करने के लिए मजबूर हूं, क्योंकि अगर सोशल मीडिया द्वारा सरकार का ध्यान खींचा जा सकता है।

আমি বাধ্য হয়ে সোশ্যাল মিডিয়াতে এই পোস্ট করছি, কারণ যদি সোশ্যাল মিডিয়ার দ্বারা সরকারের দৃষ্টি আকর্ষণ করা যায় ।

I wish I had a 'laboratory' where I could test various science data. In this hope, I have repeatedly appealed to the government. Although my article deserves to be published in a journal, I cannot publish it because it is impossible for me to bear that cost on my own initiative. Expenses in free publication journal, minimum cost, 3500 - 4000 dollars, about 250,000 - 30,000 rupees in Indian currency. For an article . But it is clear from the answer given below that there is no initiative to encourage government support or cooperation, general science practice. There is no instruction in that sector. So while the information and arguments given by me are acceptable, government officials are unable to accept them.

আমার আকাঙ্খা ছিল আমার একটি ‘ল্যবটারি’ থাকবে, সেখানে বিভিন্ন বিজ্ঞানের তথ্য পরীক্ষা নিরীক্ষা করব। সেই আশাতে সরকারের কাছে বার বার আবেদন করেছি।

আমার প্রবন্ধ জার্নালে প্রকাশের যোগ্য হলেও, প্রকাশ করতে পারছিনা কারণ, ব্যক্তিগত উদ্যোগে সেই খরচ বহন করা আমার পক্ষে অসম্ভব। ফ্রি পাবলিকেশন জার্নালে ব্যয়, সর্ব নিন্ম খরচ, ৩৫০০ – ৪০০০ ডলার, ভারতীয় মুদ্রায় প্রায় ২৫০০০০ – ৩০০০০ টাকা। একটি প্রবন্ধের জন্য।

কিন্তু নিম্নে দেওয়া উত্তর থেকে স্পষ্ট, যে সরকারী সহায়তা বা সহযোগিতা, সাধারন বিজ্ঞান চর্চাকে উৎসাহ দেবার জন্য কোন উদ্যোগ নেই। সেই খাতে কোন নির্দেশ নেই। তাই আমার দেওয়া তথ্য এবং যুক্তি গ্রহণ যোগ্য হলেও সরকারী আধিকারীরা গ্রহণ করতে অক্ষম।

६ साल तक देश के तमाम पोर्टलों के माध्यम से मेरा सवाल था: अगर मेरे विज्ञान आधारित लेख में दी गई जानकारी बेकार है, तो मुझे सीधे बताएं, अन्यथा अगर कोई स्वीकार्य तर्क है तो उसे आधिकारिक तौर पर क्यों न स्वीकार किया जाए?